Solve.
-$$4 r ^ { 2 } + 24 r = - 19$$
A) $\frac { - 24 \pm \sqrt { 17 } } { 2 }$
B) $\frac { - 6 \pm \sqrt { 55 } } { 2 }$
C) $\frac { - 6 \pm \sqrt { 17 } } { 2 }$
D) $\frac { - 6 \pm \sqrt { 17 } } { 8 }$

Question

Quizplus · Accepted Answer

The equation can be rewritten as $4r^2 + 24r + 19 = 0$. Using the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 4$, $b = 24$, and $c = 19$, we find the solutions to be $\frac{-6 \pm \sqrt{17}}{2}$.

Solve - 4r2+24r=−194 r ^ { 2 } + 24 r = - 194r2+24r=−19

Solve
- $4 r ^ { 2 } + 24 r = - 19$