Solve the equation by using the quadratic formula.
-$$6 y + 3 = - 4 y ^ { 2 }$$
A) $\left\{ - \frac { 3 } { 4 } + \frac { i \sqrt { 3 } } { 4 } , - \frac { 3 } { 4 } - \frac { i \sqrt { 3 } } { 4 } \right\}$
B) $\left\{ - \frac { 9 } { 2 } , \frac { 1 } { 3 } \right\}$
C) $\left\{ - \frac { 1 } { 4 } + \frac { \sqrt { 105 } } { 12 } , - \frac { 1 } { 4 } - \frac { \sqrt { 105 } } { 12 } \right\}$
D) $\{ 2 + \sqrt { 3 } , 2 - \sqrt { 3 } \}$

Question

Quizplus · Accepted Answer

First, rearrange the equation to standard form: $0 = -4y^2 - 6y - 3$. Then, apply the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = -4$, $b = -6$, and $c = -3$. This results in $y = \frac{6 \pm \sqrt{(-6)^2 - 4(-4)(-3)}}{2(-4)}$, simplifying to $y = \frac{6 \pm \sqrt{36 - 48}}{-8}$, which further simplifies to $y = \frac{6 \pm \sqrt{-12}}{-8}$ or $y = \frac{6 \pm 2i\sqrt{3}}{-8}$, leading to the solutions $\left\{ - \frac { 3 } { 4 } + \frac { i \sqrt { 3 } } { 4 } , - \frac { 3 } { 4 } - \frac { i \sqrt { 3 } } { 4 } \right\}$.

Solve the Equation by Using the Quadratic Formula 6y+3=−4y26 y + 3 = - 4 y ^ { 2 }6y+3=−4y2

Solve the Equation by Using the Quadratic Formula $6 y + 3 = - 4 y ^ { 2 }$