X^2+8x-9912=0

Solution for X^2+8x-9912=0 equation:

X^2+8X-9912=0

a = 1; b = 8; c = -9912;
Δ = b2-4ac
Δ = 82-4·1·(-9912)
Δ = 39712
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:

$\sqrt{\Delta}=\sqrt{39712}=\sqrt{16*2482}=\sqrt{16}*\sqrt{2482}=4\sqrt{2482}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{2482}}{2*1}=\frac{-8-4\sqrt{2482}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{2482}}{2*1}=\frac{-8+4\sqrt{2482}}{2}
$