x^2+12x-248=0

Solution for x^2+12x-248=0 equation:

x^2+12x-248=0

a = 1; b = 12; c = -248;
Δ = b2-4ac
Δ = 122-4·1·(-248)
Δ = 1136
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:

$\sqrt{\Delta}=\sqrt{1136}=\sqrt{16*71}=\sqrt{16}*\sqrt{71}=4\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{71}}{2*1}=\frac{-12-4\sqrt{71}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{71}}{2*1}=\frac{-12+4\sqrt{71}}{2}
$