Lösung 2.3:2e - Versionsgeschichte

Dagmar Timmreck um 08:28, 13. Mai 2011

Dagmar Timmreck — Fri, 13 May 2011 08:28:36 GMT

← Nächstältere Version		Version vom 08:28, 13. Mai 2011
Zeile 25:		Zeile 25:
	Schließlich kontrollieren wir unsere Antwort, indem wir kontrollieren, ob <math>x=-1</math> und <math>x=3/5</math> die ursprüngliche Gleichung erfüllen		Schließlich kontrollieren wir unsere Antwort, indem wir kontrollieren, ob <math>x=-1</math> und <math>x=3/5</math> die ursprüngliche Gleichung erfüllen

-	:*''x~~'' ~~=~~ ~~1: ~~<math>~~\ \text{Linke Seite} = 5\cdot (-1)^{2} + 2\cdot (-1) - 3 = 5 - 2 - 3 = 0 = \text{Rechte Seite,}</math>	+	:* <math>x = -1:\ \text{Linke Seite} = 5\cdot (-1)^{2} + 2\cdot (-1) - 3 = 5 - 2 - 3 = 0 = \text{Rechte Seite,}</math>

-	:*''x~~'' ~~=~~ ~~3/5: ~~<math>~~\ \text{Linke Seite} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{Rechte Seite.}</math>	+	:* <math>x = 3/5: \ \text{Linke Seite} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{Rechte Seite.}</math>

	Alternativer Lösungsweg: [[2.3:2e_alt5\|p-q_Formel]]		Alternativer Lösungsweg: [[2.3:2e_alt5\|p-q_Formel]]

Ombtut3 um 12:08, 9. Sep. 2009

Ombtut3 — Wed, 09 Sep 2009 12:08:03 GMT

← Nächstältere Version		Version vom 12:08, 9. Sep. 2009
Zeile 28:		Zeile 28:

	:*''x'' = 3/5: <math>\ \text{Linke Seite} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{Rechte Seite.}</math>		:*''x'' = 3/5: <math>\ \text{Linke Seite} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{Rechte Seite.}</math>
		+
		+	Alternativer Lösungsweg: [[2.3:2e_alt5\|p-q_Formel]]

Moni um 09:31, 9. Aug. 2009

Moni — Sun, 09 Aug 2009 09:31:36 GMT

← Nächstältere Version		Version vom 09:31, 9. Aug. 2009
Zeile 17:		Zeile 17:
	{{Abgesetzte Formel\|\|<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}</math>}}		{{Abgesetzte Formel\|\|<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}</math>}}

-	geschrieben werden, und hat ~~daher~~ die Wurzeln	+	geschrieben werden und hat die Wurzeln

-	:*<math>x+\tfrac{1}{5} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}</math> nachdem <math>\bigl(\tfrac{4}{5}\bigr)^{2} = \tfrac{16}{25}\,,</math> und wir bekommen <math>x=-\tfrac{1}{5}+\tfrac{4}{5}=\tfrac{3}{5},</math>	+	:*<math>x+\tfrac{1}{5} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5} </math> nachdem <math>\bigl(\tfrac{4}{5}\bigr)^{2} = \tfrac{16}{25}\,,</math> und wir bekommen <math>x=-\tfrac{1}{5}+\tfrac{4}{5}=\tfrac{3}{5},</math>

	:*<math>x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,,</math> und wir bekommen <math>x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}</math>		:*<math>x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,,</math> und wir bekommen <math>x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}</math>

Markus.bez um 12:31, 9. Jun. 2009

Markus.bez — Tue, 09 Jun 2009 12:31:06 GMT

← Nächstältere Version		Version vom 12:31, 9. Jun. 2009
Zeile 13:		Zeile 13:
	\end{align}</math>}}		\end{align}</math>}}

-	Die Gleichung kann daher ~~wie~~	+	Die Gleichung kann daher als

	{{Abgesetzte Formel\|\|<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}</math>}}		{{Abgesetzte Formel\|\|<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}</math>}}
Zeile 23:		Zeile 23:
	:*<math>x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,,</math> und wir bekommen <math>x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}</math>		:*<math>x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,,</math> und wir bekommen <math>x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}</math>

-	Schließlich kontrollieren wir unsere Antwort, indem wir kontrollieren ob <math>x=-1</math> und <math>x=3/5</math> die ursprüngliche Gleichung erfüllen	+	Schließlich kontrollieren wir unsere Antwort, indem wir kontrollieren, ob <math>x=-1</math> und <math>x=3/5</math> die ursprüngliche Gleichung erfüllen

	:*''x'' = 1: <math>\ \text{Linke Seite} = 5\cdot (-1)^{2} + 2\cdot (-1) - 3 = 5 - 2 - 3 = 0 = \text{Rechte Seite,}</math>		:*''x'' = 1: <math>\ \text{Linke Seite} = 5\cdot (-1)^{2} + 2\cdot (-1) - 3 = 5 - 2 - 3 = 0 = \text{Rechte Seite,}</math>

	:*''x'' = 3/5: <math>\ \text{Linke Seite} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{Rechte Seite.}</math>		:*''x'' = 3/5: <math>\ \text{Linke Seite} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{Rechte Seite.}</math>

Martin um 15:01, 16. Mär. 2009

Martin — Mon, 16 Mar 2009 15:01:52 GMT

← Nächstältere Version		Version vom 15:01, 16. Mär. 2009
Zeile 19:		Zeile 19:
	geschrieben werden, und hat daher die Wurzeln		geschrieben werden, und hat daher die Wurzeln

-	:*<math>x+\tfrac{1}{5} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}</math> ~~because~~ <math>\bigl(\tfrac{4}{5}\bigr)^{2} = \tfrac{16}{25}\,,</math> ~~which gives~~ <math>x=-\tfrac{1}{5}+\tfrac{4}{5}=\tfrac{3}{5},</math>	+	:*<math>x+\tfrac{1}{5} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}</math> nachdem <math>\bigl(\tfrac{4}{5}\bigr)^{2} = \tfrac{16}{25}\,,</math> und wir bekommen <math>x=-\tfrac{1}{5}+\tfrac{4}{5}=\tfrac{3}{5},</math>

	:*<math>x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,,</math> und wir bekommen <math>x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}</math>		:*<math>x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,,</math> und wir bekommen <math>x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}</math>

Martin um 15:00, 16. Mär. 2009

Martin — Mon, 16 Mar 2009 15:00:57 GMT

← Nächstältere Version		Version vom 15:00, 16. Mär. 2009
Zeile 1:		Zeile 1:
-	~~Write the equation in normalized form by dividing both sides by~~ 5,	+	Wir schreiben die Gleichung auf Normalform, indem wir alle Terme durch 5 dividieren,

	{{Abgesetzte Formel\|\|<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0\,\textrm{.}</math>}}

-	~~Complete the square on the left-hand side,~~	+	Wir führen die quadratische Ergänzung auf der linken Seite aus

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 13:		Zeile 13:
	\end{align}</math>}}		\end{align}</math>}}

-	~~The equation is now rewritten as~~	+	Die Gleichung kann daher wie

	{{Abgesetzte Formel\|\|<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}</math>}}		{{Abgesetzte Formel\|\|<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}</math>}}

-	~~and taking the root gives the solutions~~	+	geschrieben werden, und hat daher die Wurzeln

	:*<math>x+\tfrac{1}{5} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}</math> because <math>\bigl(\tfrac{4}{5}\bigr)^{2} = \tfrac{16}{25}\,,</math> which gives <math>x=-\tfrac{1}{5}+\tfrac{4}{5}=\tfrac{3}{5},</math>		:*<math>x+\tfrac{1}{5} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}</math> because <math>\bigl(\tfrac{4}{5}\bigr)^{2} = \tfrac{16}{25}\,,</math> which gives <math>x=-\tfrac{1}{5}+\tfrac{4}{5}=\tfrac{3}{5},</math>

-	:*<math>x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,,</math> ~~which gives~~ <math>x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}</math>	+	:*<math>x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,,</math> und wir bekommen <math>x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}</math>

-	~~Finally~~, ~~we check the answer by substituting~~ <math>x=-1</math> ~~and~~ <math>x=3/5</math> ~~into the equation:~~	+	Schließlich kontrollieren wir unsere Antwort, indem wir kontrollieren ob <math>x=-1</math> und <math>x=3/5</math> die ursprüngliche Gleichung erfüllen

-	:*''x'' = 1: <math>\ \text{~~LHS~~} = 5\cdot (-1)^{2} + 2\cdot (-1) - 3 = 5 - 2 - 3 = 0 = \text{~~RHS~~,}</math>	+	:*''x'' = 1: <math>\ \text{Linke Seite} = 5\cdot (-1)^{2} + 2\cdot (-1) - 3 = 5 - 2 - 3 = 0 = \text{Rechte Seite,}</math>

-	:*''x'' = 3/5: <math>\ \text{~~LHS~~} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{~~RHS~~.}</math>	+	:*''x'' = 3/5: <math>\ \text{Linke Seite} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{Rechte Seite.}</math>

Tekbot: hat „Solution 2.3:2e“ nach „Lösung 2.3:2e“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 14:07:45 GMT

hat „Solution 2.3:2e“ nach „Lösung 2.3:2e“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:07, 22. Okt. 2008

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

Tekbot — Wed, 22 Oct 2008 08:32:18 GMT

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 08:32, 22. Okt. 2008
Zeile 1:		Zeile 1:
	Write the equation in normalized form by dividing both sides by 5,		Write the equation in normalized form by dividing both sides by 5,

-	{{~~Displayed math~~\|\|<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0\,\textrm{.}</math>}}

	Complete the square on the left-hand side,		Complete the square on the left-hand side,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	x^{2}+\frac{2}{5}x-\frac{3}{5}		x^{2}+\frac{2}{5}x-\frac{3}{5}
	&= \Bigl(x+\frac{2/5}{2}\Bigr)^{2} - \Bigl(\frac{2/5}{2}\Bigr)^{2} - \frac{3}{5}\\[5pt]		&= \Bigl(x+\frac{2/5}{2}\Bigr)^{2} - \Bigl(\frac{2/5}{2}\Bigr)^{2} - \frac{3}{5}\\[5pt]
Zeile 15:		Zeile 15:
	The equation is now rewritten as		The equation is now rewritten as

-	{{~~Displayed math~~\|\|<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}</math>}}	+	{{Abgesetzte Formel\|\|<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}</math>}}

	and taking the root gives the solutions		and taking the root gives the solutions

Tek um 08:13, 29. Sep. 2008

Tek — Mon, 29 Sep 2008 08:13:15 GMT

← Nächstältere Version		Version vom 08:13, 29. Sep. 2008
Zeile 1:		Zeile 1:
-	Write the equation in normalized form by dividing both sides by	+	Write the equation in normalized form by dividing both sides by 5,
-	~~<math>~~5~~</math>~~,	+

		+	{{Displayed math\|\|<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0\,\textrm{.}</math>}}

-	~~<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0</math>~~	+	Complete the square on the left-hand side,
-		+
-		+
-	~~Completing~~ the square on the left-hand side,	+
-	~~<math></math>~~	+
-		+
-	~~<math></math>~~	+
-		+
-	~~<math></math>~~	+
-		+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& x^{2}+\frac{2}{5}x-\frac{3}{5}=\left( x+\frac{\frac{2}{5}}{2} \right)^{2}-\left( \frac{\frac{2}{5}}{2} \right)^{2}-\frac{3}{5} \\~~	+
-	~~& =\left( x+\frac{1}{5} \right)^{2}-\left( \frac{1}{5} \right)^{2}-\frac{3}{5}=\left( x+\frac{1}{5} \right)^{2}-\frac{1}{25}-\frac{3\centerdot 5}{25}=\left( x+\frac{1}{5} \right)^{2}-\frac{16}{25} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	x^{2}+\frac{2}{5}x-\frac{3}{5}
		+	&= \Bigl(x+\frac{2/5}{2}\Bigr)^{2} - \Bigl(\frac{2/5}{2}\Bigr)^{2} - \frac{3}{5}\\[5pt]
		+	&= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \Bigl(\frac{1}{5}\Bigr)^{2} - \frac{3}{5}\\[5pt]
		+	&= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \frac{1}{25} - \frac{3\cdot 5}{25}\\[5pt]
		+	&= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \frac{16}{25}\,\textrm{.}
		+	\end{align}</math>}}

	The equation is now rewritten as		The equation is now rewritten as

-		+	{{Displayed math\|\|<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}</math>}}
-	<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}</math>	+
-		+

	and taking the root gives the solutions		and taking the root gives the solutions

		+	:*<math>x+\tfrac{1}{5} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}</math> because <math>\bigl(\tfrac{4}{5}\bigr)^{2} = \tfrac{16}{25}\,,</math> which gives <math>x=-\tfrac{1}{5}+\tfrac{4}{5}=\tfrac{3}{5},</math>

-	<math>x+\~~frac~~{1}{5}=\sqrt{\~~frac~~{16}{25}}~~=\frac{4}{5}</math>~~	+	:*<math>x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,,</math> which gives <math>x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}</math>
-	~~because~~	+
-	~~<math>\left( \frac{4}{5}^{2} \right)=\frac{16}{25}</math>~~	+
-	~~which gives~~	+
-	~~<math>x~~=-\~~frac{1}{5}+\frac~~{4}{5}=\~~frac{3}{5}~~,</math>	+
-		+
-		+
-	~~<math>x+\frac{1}{5}=-\sqrt{\frac{16}{25}}=-\frac{4}{5}</math>~~	+
-	which gives	+
-	<math>x=-\~~frac~~{1}{5}-\~~frac~~{4}{5}=-1,~~</math>~~	+
-		+
-		+
-	~~Finally, we check the answer by substituting~~	+
-	~~<math>x=-~~\~~text~~{1 }</math>	+
-	~~and~~	+
-	~~<math>x=\text{3}/\text{5 }</math>~~	+
-	~~into the equation:~~	+

		+	Finally, we check the answer by substituting <math>x=-1</math> and <math>x=3/5</math> into the equation:

-	<math>~~x=-~~\text{1 }~~</math>: LHS~~	+	:*''x'' = 1: <math>\ \text{LHS} = 5\cdot (-1)^{2} + 2\cdot (-1) - 3 = 5 - 2 - 3 = 0 = \text{RHS,}</math>
-	~~<math>~~=5\~~centerdot \left~~( -1 ~~\right~~)^{2}+2\~~centerdot \left~~( -1 ~~\right~~)-3=5-2-3=0=</math>	+
-	~~RHS~~	+

-	~~<math>~~x=~~\text{~~3}/~~\text{~~5 }</math>: LHS	+	:*''x'' = 3/5: <math>\ \text{LHS} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{RHS.}</math>
-	~~<math>~~=5\~~centerdot~~ \~~left~~( \~~frac~~{3}{5} \~~right~~)^{2}+2\~~centerdot~~ \~~left~~( \~~frac~~{3}{5} \~~right~~)-3=5\~~centerdot~~ \~~frac~~{9}{25}+\~~frac~~{6}{5}-\~~frac~~{3\~~centerdot~~ 5}{5}=\~~frac~~{~~9+6-15~~}~~{5}=0=~~</math>	+
-	~~RHS~~	+

Ian um 14:17, 20. Sep. 2008

Ian — Sat, 20 Sep 2008 14:17:04 GMT

← Nächstältere Version		Version vom 14:17, 20. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	Write the equation in normalized form by dividing both sides by
-	<~~center~~> ~~[[Image~~:~~2_3_2e.gif]]~~ </~~center~~>	+	<math>5</math>,
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0</math>
		+
		+
		+	Completing the square on the left-hand side,
		+	<math></math>
		+
		+	<math></math>
		+
		+	<math></math>
		+
		+
		+
		+	<math>\begin{align}
		+	& x^{2}+\frac{2}{5}x-\frac{3}{5}=\left( x+\frac{\frac{2}{5}}{2} \right)^{2}-\left( \frac{\frac{2}{5}}{2} \right)^{2}-\frac{3}{5} \\
		+	& =\left( x+\frac{1}{5} \right)^{2}-\left( \frac{1}{5} \right)^{2}-\frac{3}{5}=\left( x+\frac{1}{5} \right)^{2}-\frac{1}{25}-\frac{3\centerdot 5}{25}=\left( x+\frac{1}{5} \right)^{2}-\frac{16}{25} \\
		+	\end{align}</math>
		+
		+
		+	The equation is now rewritten as
		+
		+
		+	<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}</math>
		+
		+
		+	and taking the root gives the solutions
		+
		+
		+	<math>x+\frac{1}{5}=\sqrt{\frac{16}{25}}=\frac{4}{5}</math>
		+	because
		+	<math>\left( \frac{4}{5}^{2} \right)=\frac{16}{25}</math>
		+	which gives
		+	<math>x=-\frac{1}{5}+\frac{4}{5}=\frac{3}{5},</math>
		+
		+
		+	<math>x+\frac{1}{5}=-\sqrt{\frac{16}{25}}=-\frac{4}{5}</math>
		+	which gives
		+	<math>x=-\frac{1}{5}-\frac{4}{5}=-1,</math>
		+
		+
		+	Finally, we check the answer by substituting
		+	<math>x=-\text{1 }</math>
		+	and
		+	<math>x=\text{3}/\text{5 }</math>
		+	into the equation:
		+
		+
		+	<math>x=-\text{1 }</math>: LHS
		+	<math>=5\centerdot \left( -1 \right)^{2}+2\centerdot \left( -1 \right)-3=5-2-3=0=</math>
		+	RHS
		+
		+	<math>x=\text{3}/\text{5 }</math>: LHS
		+	<math>=5\centerdot \left( \frac{3}{5} \right)^{2}+2\centerdot \left( \frac{3}{5} \right)-3=5\centerdot \frac{9}{25}+\frac{6}{5}-\frac{3\centerdot 5}{5}=\frac{9+6-15}{5}=0=</math>
		+	RHS