a^2+32a-128=0

Solution for a^2+32a-128=0 equation:

a^2+32a-128=0

a = 1; b = 32; c = -128;
Δ = b2-4ac
Δ = 322-4·1·(-128)
Δ = 1536
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{6}}{2*1}=\frac{-32-16\sqrt{6}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{6}}{2*1}=\frac{-32+16\sqrt{6}}{2}
$