X^2+10x-480=0

Solution for X^2+10x-480=0 equation:

X^2+10X-480=0

a = 1; b = 10; c = -480;
Δ = b2-4ac
Δ = 102-4·1·(-480)
Δ = 2020
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:

$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{505}}{2*1}=\frac{-10-2\sqrt{505}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{505}}{2*1}=\frac{-10+2\sqrt{505}}{2}
$