2x2+20x=-38

Solution for 2x2+20x=-38 equation:

2x^2+20x=-38

We move all terms to the left:

2x^2+20x-(-38)=0

We add all the numbers together, and all the variables

2x^2+20x+38=0

a = 2; b = 20; c = +38;
Δ = b2-4ac
Δ = 202-4·2·38
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{6}}{2*2}=\frac{-20-4\sqrt{6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{6}}{2*2}=\frac{-20+4\sqrt{6}}{4}
$