x^2+20x-1728=0

Solution for x^2+20x-1728=0 equation:

x^2+20x-1728=0

a = 1; b = 20; c = -1728;
Δ = b2-4ac
Δ = 202-4·1·(-1728)
Δ = 7312
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:

$\sqrt{\Delta}=\sqrt{7312}=\sqrt{16*457}=\sqrt{16}*\sqrt{457}=4\sqrt{457}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{457}}{2*1}=\frac{-20-4\sqrt{457}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{457}}{2*1}=\frac{-20+4\sqrt{457}}{2}
$