x^2+12x-162=0

Solution for x^2+12x-162=0 equation:

x^2+12x-162=0

a = 1; b = 12; c = -162;
Δ = b2-4ac
Δ = 122-4·1·(-162)
Δ = 792
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:

$\sqrt{\Delta}=\sqrt{792}=\sqrt{36*22}=\sqrt{36}*\sqrt{22}=6\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{22}}{2*1}=\frac{-12-6\sqrt{22}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{22}}{2*1}=\frac{-12+6\sqrt{22}}{2}
$