Solution 2.3:2e
From Förberedande kurs i matematik 1
Write the equation in normalized form by dividing both sides by \displaystyle 5,
\displaystyle x^{2}+\frac{2}{5}x-\frac{3}{5}=0
Completing the square on the left-hand side,
\displaystyle
\displaystyle
\displaystyle
\displaystyle \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}
The equation is now rewritten as
\displaystyle \left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}
and taking the root gives the solutions
\displaystyle x+\frac{1}{5}=\sqrt{\frac{16}{25}}=\frac{4}{5}
because
\displaystyle \left( \frac{4}{5}^{2} \right)=\frac{16}{25}
which gives
\displaystyle x=-\frac{1}{5}+\frac{4}{5}=\frac{3}{5},
\displaystyle x+\frac{1}{5}=-\sqrt{\frac{16}{25}}=-\frac{4}{5}
which gives
\displaystyle x=-\frac{1}{5}-\frac{4}{5}=-1,
Finally, we check the answer by substituting
\displaystyle x=-\text{1 }
and
\displaystyle x=\text{3}/\text{5 }
into the equation:
\displaystyle x=-\text{1 }: LHS
\displaystyle =5\centerdot \left( -1 \right)^{2}+2\centerdot \left( -1 \right)-3=5-2-3=0=
RHS
\displaystyle x=\text{3}/\text{5 }: LHS \displaystyle =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= RHS
