2x^2+4x+20=100

Solution for 2x^2+4x+20=100 equation:

2x^2+4x+20=100

We move all terms to the left:

2x^2+4x+20-(100)=0

We add all the numbers together, and all the variables

2x^2+4x-80=0

a = 2; b = 4; c = -80;
Δ = b2-4ac
Δ = 42-4·2·(-80)
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{41}}{2*2}=\frac{-4-4\sqrt{41}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{41}}{2*2}=\frac{-4+4\sqrt{41}}{4}
$