2x^2+18x+20=-16

Solution for 2x^2+18x+20=-16 equation:

2x^2+18x+20=-16

We move all terms to the left:

2x^2+18x+20-(-16)=0

We add all the numbers together, and all the variables

2x^2+18x+36=0

a = 2; b = 18; c = +36;
Δ = b2-4ac
Δ = 182-4·2·36
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*2}=\frac{-24}{4}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*2}=\frac{-12}{4}
=-3
$