a^2+24a=-80

Solution for a^2+24a=-80 equation:

a^2+24a=-80

We move all terms to the left:

a^2+24a-(-80)=0

We add all the numbers together, and all the variables

a^2+24a+80=0

a = 1; b = 24; c = +80;
Δ = b2-4ac
Δ = 242-4·1·80
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16}{2*1}=\frac{-40}{2}
=-20
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16}{2*1}=\frac{-8}{2}
=-4
$