x^2+12x+32=128

Solution for x^2+12x+32=128 equation:

x^2+12x+32=128

We move all terms to the left:

x^2+12x+32-(128)=0

We add all the numbers together, and all the variables

x^2+12x-96=0

a = 1; b = 12; c = -96;
Δ = b2-4ac
Δ = 122-4·1·(-96)
Δ = 528
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{33}}{2*1}=\frac{-12-4\sqrt{33}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{33}}{2*1}=\frac{-12+4\sqrt{33}}{2}
$