0=x^2+2x-160

Solution for 0=x^2+2x-160 equation:

0=x^2+2x-160

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-x^2-2x+160=0

We add all the numbers together, and all the variables

-1x^2-2x+160=0

a = -1; b = -2; c = +160;
Δ = b2-4ac
Δ = -22-4·(-1)·160
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{161}}{2*-1}=\frac{2-2\sqrt{161}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{161}}{2*-1}=\frac{2+2\sqrt{161}}{-2}
$