## Algebra 2 (1st Edition)

$$x=1,\:x=-4$$
Completing the square to solve, we find: $$3\left(x^2+3x-4\right) \\ 3\left(x^2+3x-4+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\right) \\ 3\left(x+\frac{3}{2}\right)^2-\frac{75}{4} \\ \left(x+\frac{3}{2}\right)^2=\frac{25}{4} \\ x=1,\:x=-4$$