x^2+8x-160=x

Solution for x^2+8x-160=x equation:

x^2+8x-160=x

We move all terms to the left:

x^2+8x-160-(x)=0

We add all the numbers together, and all the variables

x^2+7x-160=0

a = 1; b = 7; c = -160;
Δ = b2-4ac
Δ = 72-4·1·(-160)
Δ = 689
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{689}}{2*1}=\frac{-7-\sqrt{689}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{689}}{2*1}=\frac{-7+\sqrt{689}}{2}
$