# How do you solve r^2+8r+5=0 using the quadratic formula?

Oct 21, 2015

Think of the coefficient of ${r}^{2}$ as 1, so really it is $1 {r}^{2} + 8 r + 5$, now you just plug in those numbers into the quadratic formula. It would be
$\frac{- 8 + \sqrt{{8}^{2} - \left(4\right) \left(1\right) \left(5\right)}}{2 \cdot 1} = - 0.68$
$\frac{- 8 - \sqrt{{8}^{2} - \left(4\right) \left(1\right) \left(5\right)}}{2 \cdot 1} = - 7.32$
$x = - 0.68 , - 7.32$