x^2+4x-800=0

Solution for x^2+4x-800=0 equation:

x^2+4x-800=0

a = 1; b = 4; c = -800;
Δ = b2-4ac
Δ = 42-4·1·(-800)
Δ = 3216
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:

$\sqrt{\Delta}=\sqrt{3216}=\sqrt{16*201}=\sqrt{16}*\sqrt{201}=4\sqrt{201}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{201}}{2*1}=\frac{-4-4\sqrt{201}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{201}}{2*1}=\frac{-4+4\sqrt{201}}{2}
$