x^2+20x+28=0

Solution for x^2+20x+28=0 equation:

x^2+20x+28=0

a = 1; b = 20; c = +28;
Δ = b2-4ac
Δ = 202-4·1·28
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{2}}{2*1}=\frac{-20-12\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{2}}{2*1}=\frac{-20+12\sqrt{2}}{2}
$