## Algebra 2 (1st Edition)

$$x=-3\pm\sqrt3$$
The student should not have used the quadratic formula until the solution was set equal to zero. Thus, we first put the function in the form $ax^2+bx+c$. Then, using the quadratic formula, we obtain: $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ $$x=\frac{-6\pm \sqrt{(6)^2-4(1)(6)}}{2(1)}$$ $$x=-3\pm\sqrt3$$