2x^2+20x+100=108^2

Solution for 2x^2+20x+100=108^2 equation:

2x^2+20x+100=108^2

We move all terms to the left:

2x^2+20x+100-(108^2)=0

We add all the numbers together, and all the variables

2x^2+20x-11564=0

a = 2; b = 20; c = -11564;
Δ = b2-4ac
Δ = 202-4·2·(-11564)
Δ = 92912
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:

$\sqrt{\Delta}=\sqrt{92912}=\sqrt{16*5807}=\sqrt{16}*\sqrt{5807}=4\sqrt{5807}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{5807}}{2*2}=\frac{-20-4\sqrt{5807}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{5807}}{2*2}=\frac{-20+4\sqrt{5807}}{4}
$