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} (x+3)^{2}-(x-5)^{2} &= (x^{2}+2\cdot 3x+3^{2})-(x^{2}-2\cdot 5x+5^{2})\\[5pt] &= x^{2}+6x+9-x^{2}+10x-25\\[5pt] &=16x-16\,\textrm{.} \end{align}

Thus, the equation is

\displaystyle 16x-16=6x+4\,\textrm{.}

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

\displaystyle \begin{align} 16x-6x&=4+16\,,\\[5pt] 10x&=20\,\textrm{.} \end{align}

Divide both sides by 10 to get the answer

\displaystyle x=\frac{20}{10}=2\,\textrm{.}

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

\displaystyle \begin{align} \text{LHS} &= (2+3)^{2}-(2-5)^{2} = 5^{2}-(-3)^{2} = 25-9 = 16,\\[5pt] \text{RHS} &= 6\cdot 2+4 = 12+4 = 16\,\textrm{.} \end{align}