-3r^2-12=-33

Solution for -3r^2-12=-33 equation:

-3r^2-12=-33

We move all terms to the left:

-3r^2-12-(-33)=0

We add all the numbers together, and all the variables

-3r^2+21=0

a = -3; b = 0; c = +21;
Δ = b2-4ac
Δ = 02-4·(-3)·21
Δ = 252
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{7}}{2*-3}=\frac{0-6\sqrt{7}}{-6}
=-\frac{6\sqrt{7}}{-6}
=-\frac{\sqrt{7}}{-1}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{7}}{2*-3}=\frac{0+6\sqrt{7}}{-6}
=\frac{6\sqrt{7}}{-6}
=\frac{\sqrt{7}}{-1}
$