-3x^2+36x=-90

Solution for -3x^2+36x=-90 equation:

-3x^2+36x=-90

We move all terms to the left:

-3x^2+36x-(-90)=0

We add all the numbers together, and all the variables

-3x^2+36x+90=0

a = -3; b = 36; c = +90;
Δ = b2-4ac
Δ = 362-4·(-3)·90
Δ = 2376
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2376}=\sqrt{36*66}=\sqrt{36}*\sqrt{66}=6\sqrt{66}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-6\sqrt{66}}{2*-3}=\frac{-36-6\sqrt{66}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+6\sqrt{66}}{2*-3}=\frac{-36+6\sqrt{66}}{-6}
$