-3x^2+12=-30

Solution for -3x^2+12=-30 equation:

-3x^2+12=-30

We move all terms to the left:

-3x^2+12-(-30)=0

We add all the numbers together, and all the variables

-3x^2+42=0

a = -3; b = 0; c = +42;
Δ = b2-4ac
Δ = 02-4·(-3)·42
Δ = 504
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:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{14}}{2*-3}=\frac{0-6\sqrt{14}}{-6}
=-\frac{6\sqrt{14}}{-6}
=-\frac{\sqrt{14}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{14}}{2*-3}=\frac{0+6\sqrt{14}}{-6}
=\frac{6\sqrt{14}}{-6}
=\frac{\sqrt{14}}{-1}
$