x^2+8x-390=0

Solution for x^2+8x-390=0 equation:

x^2+8x-390=0

a = 1; b = 8; c = -390;
Δ = b2-4ac
Δ = 82-4·1·(-390)
Δ = 1624
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:

$\sqrt{\Delta}=\sqrt{1624}=\sqrt{4*406}=\sqrt{4}*\sqrt{406}=2\sqrt{406}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{406}}{2*1}=\frac{-8-2\sqrt{406}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{406}}{2*1}=\frac{-8+2\sqrt{406}}{2}
$