x^2+20x-441=0

Solution for x^2+20x-441=0 equation:

x^2+20x-441=0

a = 1; b = 20; c = -441;
Δ = b2-4ac
Δ = 202-4·1·(-441)
Δ = 2164
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:

$\sqrt{\Delta}=\sqrt{2164}=\sqrt{4*541}=\sqrt{4}*\sqrt{541}=2\sqrt{541}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{541}}{2*1}=\frac{-20-2\sqrt{541}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{541}}{2*1}=\frac{-20+2\sqrt{541}}{2}
$