8x^2+8x=160

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

8x^2+8x=160

We move all terms to the left:

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

a = 8; b = 8; c = -160;
Δ = b2-4ac
Δ = 82-4·8·(-160)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-72}{2*8}=\frac{-80}{16}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+72}{2*8}=\frac{64}{16}
=4
$