x^2+12x-968=0

Solution for x^2+12x-968=0 equation:

x^2+12x-968=0

a = 1; b = 12; c = -968;
Δ = b2-4ac
Δ = 122-4·1·(-968)
Δ = 4016
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:

$\sqrt{\Delta}=\sqrt{4016}=\sqrt{16*251}=\sqrt{16}*\sqrt{251}=4\sqrt{251}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{251}}{2*1}=\frac{-12-4\sqrt{251}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{251}}{2*1}=\frac{-12+4\sqrt{251}}{2}
$