a^2+24a-256=0

Solution for a^2+24a-256=0 equation:

a^2+24a-256=0

a = 1; b = 24; c = -256;
Δ = b2-4ac
Δ = 242-4·1·(-256)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-40}{2*1}=\frac{-64}{2}
=-32
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+40}{2*1}=\frac{16}{2}
=8
$