-3x^2+40=-20

Solution for -3x^2+40=-20 equation:

-3x^2+40=-20

We move all terms to the left:

-3x^2+40-(-20)=0

We add all the numbers together, and all the variables

-3x^2+60=0

a = -3; b = 0; c = +60;
Δ = b2-4ac
Δ = 02-4·(-3)·60
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{5}}{2*-3}=\frac{0-12\sqrt{5}}{-6}
=-\frac{12\sqrt{5}}{-6}
=-\frac{2\sqrt{5}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{5}}{2*-3}=\frac{0+12\sqrt{5}}{-6}
=\frac{12\sqrt{5}}{-6}
=\frac{2\sqrt{5}}{-1}
$