How do you solve 6r^2+5r=-1 using the quadratic formula?

Apr 17, 2017

$- \frac{6}{12}$ and $- \frac{4}{12}$

Explanation:

Compute $\Delta$ first. $\Delta = {b}^{2} - \left(4 a c\right)$ where a=6, b=5 and c=1 in the arranged equation: $6 {r}^{2} + 5 r + 1 = 0$

Now $\Delta = 1 = {5}^{2} - \left(4 \cdot 6 \cdot 1\right)$

$\sqrt{\Delta} = 1$

Now compute rs based on $r = \frac{- b - \sqrt{\Delta}}{2 a}$ and $r = \frac{- b + \sqrt{\Delta}}{2 a}$

$r 1 = \frac{- 5 - 1}{2 \cdot 6} = - \frac{6}{12}$ and $r 2 = \frac{- 5 + 1}{2 \cdot 6} = - \frac{4}{12}$

These are the answers: $- \frac{6}{12}$ and $- \frac{4}{12}$