2x^2+10x+20=-10x-4

Solution for 2x^2+10x+20=-10x-4 equation:

2x^2+10x+20=-10x-4

We move all terms to the left:

2x^2+10x+20-(-10x-4)=0

We get rid of parentheses

2x^2+10x+10x+4+20=0

We add all the numbers together, and all the variables

2x^2+20x+24=0

a = 2; b = 20; c = +24;
Δ = b2-4ac
Δ = 202-4·2·24
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{13}}{2*2}=\frac{-20-4\sqrt{13}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{13}}{2*2}=\frac{-20+4\sqrt{13}}{4}
$