## Algebra 2 (1st Edition)

$$x=2\sqrt{6}+3,\:x=-2\sqrt{6}+3$$
We complete the square to find: $$x^2-6x-15+\left(-3\right)^2-\left(-3\right)^2 \\ \left(x-3\right)^2-15-\left(-3\right)^2 \\ \left(x-3\right)^2-24 \\ \left(x-3\right)^2=24 \\ x=2\sqrt{6}+3,\:x=-2\sqrt{6}+3$$