2(x^2+20x+99)=0

Solution for 2(x^2+20x+99)=0 equation:

2(x^2+20x+99)=0

We multiply parentheses

2x^2+40x+198=0

a = 2; b = 40; c = +198;
Δ = b2-4ac
Δ = 402-4·2·198
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4}{2*2}=\frac{-44}{4}
=-11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4}{2*2}=\frac{-36}{4}
=-9
$