From Förberedande kurs i matematik 1

We can simplify the left-hand side in the equation by expanding the squares using the squaring rule:

\displaystyle \begin{align} & \left( x+3 \right)^{2}-\left( x-5 \right)^{2}=\left( x^{2}+2\centerdot 3x+3^{2} \right)-\left( x^{2}-2\centerdot 5x+5^{2} \right) \\ & =x^{2}+6x+9-x^{2}+10x-25=16x-16 \\ \end{align}

Thus, the equation is

\displaystyle 16x-16=6x+4

Now, move all " \displaystyle x "s to the left-hand side (subtract \displaystyle 6x from both sides) and the constants to the right-hand side (add \displaystyle 16 to both sides)

\displaystyle \begin{align} & 16x-6x=4+16 \\ & 10x=20 \\ \end{align}

Divide both sides by \displaystyle 10 to get the answer

\displaystyle x=\frac{20}{10}=2

Finally, we check that \displaystyle x=2 satisfies the equation in the exercise:

LHS = \displaystyle \left( 2+3 \right)^{2}-\left( 2-5 \right)^{2}=5^{2}-\left( -3 \right)^{2}=25-9=16

RHS = \displaystyle 6\centerdot 2+4=12+4=16