2x^2+4x+20=180

Solution for 2x^2+4x+20=180 equation:

2x^2+4x+20=180

We move all terms to the left:

2x^2+4x+20-(180)=0

We add all the numbers together, and all the variables

2x^2+4x-160=0

a = 2; b = 4; c = -160;
Δ = b2-4ac
Δ = 42-4·2·(-160)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-36}{2*2}=\frac{-40}{4}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+36}{2*2}=\frac{32}{4}
=8
$