Solve.
-$41 m ^ { 2 } + 100 m + 53 = 0$
A) $\frac { - 50 \pm \sqrt { 4673 } } { 41 }$
B) $\frac { - 50 \pm \sqrt { 327 } } { 82 }$
C) $\frac { - 100 \pm \sqrt { 327 } } { 41 }$
D) $\frac { - 50 \pm \sqrt { 327 } } { 41 }$

Question

Quizplus · Accepted Answer

The quadratic formula is $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 41$, $b = 100$, and $c = 53$. Plugging these values in, we get $\frac{-100 \pm \sqrt{100^2 - 4(41)(53)}}{2(41)}$, which simplifies to $\frac{-100 \pm \sqrt{10000 - 8624}}{82}$, further simplifying to $\frac{-100 \pm \sqrt{1376}}{82}$, and $\sqrt{1376} = \sqrt{327}$. Thus, the correct answer is $\frac{-50 \pm \sqrt{327}}{41}$.

Solve - 41m2+100m+53=041 m ^ { 2 } + 100 m + 53 = 041m2+100m+53=0

Solve
- $41 m ^ { 2 } + 100 m + 53 = 0$