x^2+20x+54=-10

Solution for x^2+20x+54=-10 equation:

x^2+20x+54=-10

We move all terms to the left:

x^2+20x+54-(-10)=0

We add all the numbers together, and all the variables

x^2+20x+64=0

a = 1; b = 20; c = +64;
Δ = b2-4ac
Δ = 202-4·1·64
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12}{2*1}=\frac{-32}{2}
=-16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12}{2*1}=\frac{-8}{2}
=-4
$