x^2+20x=160

Solution for x^2+20x=160 equation:

x^2+20x=160

We move all terms to the left:

x^2+20x-(160)=0

a = 1; b = 20; c = -160;
Δ = b2-4ac
Δ = 202-4·1·(-160)
Δ = 1040
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:

$\sqrt{\Delta}=\sqrt{1040}=\sqrt{16*65}=\sqrt{16}*\sqrt{65}=4\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{65}}{2*1}=\frac{-20-4\sqrt{65}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{65}}{2*1}=\frac{-20+4\sqrt{65}}{2}
$