x^2+20x-62=0

Solution for x^2+20x-62=0 equation:

x^2+20x-62=0

a = 1; b = 20; c = -62;
Δ = b2-4ac
Δ = 202-4·1·(-62)
Δ = 648
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:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-18\sqrt{2}}{2*1}=\frac{-20-18\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+18\sqrt{2}}{2*1}=\frac{-20+18\sqrt{2}}{2}
$