10x+x^2-640=0

Solution for 10x+x^2-640=0 equation:

10x+x^2-640=0

a = 1; b = 10; c = -640;
Δ = b2-4ac
Δ = 102-4·1·(-640)
Δ = 2660
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:

$\sqrt{\Delta}=\sqrt{2660}=\sqrt{4*665}=\sqrt{4}*\sqrt{665}=2\sqrt{665}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{665}}{2*1}=\frac{-10-2\sqrt{665}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{665}}{2*1}=\frac{-10+2\sqrt{665}}{2}
$