2x^2+2x-4=20

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

2x^2+2x-4=20

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x^2+2x-24=0

a = 2; b = 2; c = -24;
Δ = b2-4ac
Δ = 22-4·2·(-24)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*2}=\frac{12}{4}
=3
$