-3r2=9r+6

Solution for -3r2=9r+6 equation:

-3r^2=9r+6

We move all terms to the left:

-3r^2-(9r+6)=0

We get rid of parentheses

-3r^2-9r-6=0

a = -3; b = -9; c = -6;
Δ = b2-4ac
Δ = -92-4·(-3)·(-6)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3}{2*-3}=\frac{6}{-6}
=-1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3}{2*-3}=\frac{12}{-6}
=-2
$