Solve by completing the square.
-$$y ^ { 2 } = - 3 y + 9$$
A) $- 3 \pm 3 \sqrt { 5 }$
B) $\frac { - 3 - 3 \sqrt { 5 } } { 2 }$
C) $\frac { - 3 \pm 3 \sqrt { 5 } } { 2 }$
D) $\frac { 3 + 3 \sqrt { 5 } } { 2 }$

Question

Quizplus · Accepted Answer

First, we move all the terms to one side to get $y^2+3y-9=0$. Then we complete the square by adding $\left(\frac{3}{2}\right)^2=\frac{9}{4}$ to both sides of the equation:
$$ y^2 + 3y + \frac{9}{4} = \frac{45}{4} $$
This can be written as:
$$ \left(y+\frac{3}{2}\right)^2 = \frac{45}{4} $$
Taking the square root of both sides, we get:
$$ y+\frac{3}{2} = \pm \frac{3\sqrt{5}}{2} $$
Solving for $y$ gives us:
$$ y=\frac{-3\pm3\sqrt{5}}{2} $$
Therefore, the answer is $\boxed{\textbf{(C)}}$.

Solve by Completing the Square y2=−3y+9y ^ { 2 } = - 3 y + 9y2=−3y+9

Solve by Completing the Square $y ^ { 2 } = - 3 y + 9$