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

Question

1864 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-17T10:10:05

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

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

Now $Delta=1=5^2-(4*6*1)$

$sqrt Delta=1$

Now compute rs based on $r=(-b-sqrtDelta)/(2a)$ and $r=(-b+sqrtDelta)/(2a)$

$r1=(-5-1)/(2*6)=-6/12$ and $r2=(-5+1)/(2*6)=-4/12$

These are the answers: $-6/12$ and $-4/12$

How do you solve 6r^2+5r=-16r2+5r=−16r^2+5r=-1 using the quadratic formula?