x2+20x-12=0

Solution for x2+20x-12=0 equation:

x2+20x-12=0

We add all the numbers together, and all the variables

x^2+20x-12=0

a = 1; b = 20; c = -12;
Δ = b2-4ac
Δ = 202-4·1·(-12)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{7}}{2*1}=\frac{-20-8\sqrt{7}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{7}}{2*1}=\frac{-20+8\sqrt{7}}{2}
$