# How do you solve x^2+x-8=4?

${x}^{2} + x - 8 - 4 = 0$
${x}^{2} + x - 12 = 0$
${x}^{2} + 4 x - 3 x - 12 = 0$
$x \left(x + 4\right) - 3 \left(x + 4\right) = 0$
$\left(x + 4\right) \cdot \left(x - 3\right) = 0$
$x + 4 = 0 , x - 3 = 0$
$x = - 4 , x = 3$