X^2+20x+68=0

Solution for X^2+20x+68=0 equation:

X^2+20X+68=0

a = 1; b = 20; c = +68;
Δ = b2-4ac
Δ = 202-4·1·68
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{2}}{2*1}=\frac{-20-8\sqrt{2}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{2}}{2*1}=\frac{-20+8\sqrt{2}}{2}
$