20x^2+20=180

Solution for 20x^2+20=180 equation:

20x^2+20=180

We move all terms to the left:

20x^2+20-(180)=0

We add all the numbers together, and all the variables

20x^2-160=0

a = 20; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·20·(-160)
Δ = 12800
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:

$\sqrt{\Delta}=\sqrt{12800}=\sqrt{6400*2}=\sqrt{6400}*\sqrt{2}=80\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80\sqrt{2}}{2*20}=\frac{0-80\sqrt{2}}{40}
=-\frac{80\sqrt{2}}{40}
=-2\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80\sqrt{2}}{2*20}=\frac{0+80\sqrt{2}}{40}
=\frac{80\sqrt{2}}{40}
=2\sqrt{2}
$