x^2+8x-5040=0

Solution for x^2+8x-5040=0 equation:

x^2+8x-5040=0

a = 1; b = 8; c = -5040;
Δ = b2-4ac
Δ = 82-4·1·(-5040)
Δ = 20224
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:

$\sqrt{\Delta}=\sqrt{20224}=\sqrt{256*79}=\sqrt{256}*\sqrt{79}=16\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16\sqrt{79}}{2*1}=\frac{-8-16\sqrt{79}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16\sqrt{79}}{2*1}=\frac{-8+16\sqrt{79}}{2}
$