x^2+20x-4865=0

Solution for x^2+20x-4865=0 equation:

x^2+20x-4865=0

a = 1; b = 20; c = -4865;
Δ = b2-4ac
Δ = 202-4·1·(-4865)
Δ = 19860
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:

$\sqrt{\Delta}=\sqrt{19860}=\sqrt{4*4965}=\sqrt{4}*\sqrt{4965}=2\sqrt{4965}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{4965}}{2*1}=\frac{-20-2\sqrt{4965}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{4965}}{2*1}=\frac{-20+2\sqrt{4965}}{2}
$