Question

solve for y.
$5y^{2}+41y + 8 = 0$
write each solution as an integer, proper fraction, or improper fraction in simplest form. if there are multiple solutions, separate them with commas.
$y = $
submit

Explanation:

Step1: Identify quadratic coefficients

For $5y^2 + 41y + 8 = 0$, we have $a=5$, $b=41$, $c=8$.

Step2: Apply quadratic formula

Quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Substitute values:

$$ y = \frac{-41 \pm \sqrt{41^2 - 4(5)(8)}}{2(5)} $$

Step3: Calculate discriminant

Compute $41^2 - 4(5)(8) = 1681 - 160 = 1521$, and $\sqrt{1521}=39$.

Step4: Solve for both roots

First root (using minus):

$$ y = \frac{-41 - 39}{10} = \frac{-80}{10} = -8 $$

Second root (using plus):

$$ y = \frac{-41 + 39}{10} = \frac{-2}{10} = -\frac{1}{5} $$

Answer:

$-8, -\frac{1}{5}$