# How do you find the solution set for x² + 10x + 25 = 64?

${x}^{2} + 10 x - 39 = 0$
${x}^{2} + 13 x - 3 x - 39 = 0$
$x \left(x + 13\right) - 3 \left(x + 13\right) = 0$
$\implies \left(x + 13\right) \left(x - 3\right) = 0$
$x = - 13 , 3$