x^2+12x-432=0

Solution for x^2+12x-432=0 equation:

x^2+12x-432=0

a = 1; b = 12; c = -432;
Δ = b2-4ac
Δ = 122-4·1·(-432)
Δ = 1872
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:

$\sqrt{\Delta}=\sqrt{1872}=\sqrt{144*13}=\sqrt{144}*\sqrt{13}=12\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{13}}{2*1}=\frac{-12-12\sqrt{13}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{13}}{2*1}=\frac{-12+12\sqrt{13}}{2}
$