160x+160x+640=(8x^2)+72x

Solution for 160x+160x+640=(8x^2)+72x equation:

160x+160x+640=(8x^2)+72x

We move all terms to the left:

160x+160x+640-((8x^2)+72x)=0

We add all the numbers together, and all the variables

320x-(8x^2+72x)+640=0

We get rid of parentheses

-8x^2+320x-72x+640=0

We add all the numbers together, and all the variables

-8x^2+248x+640=0

a = -8; b = 248; c = +640;
Δ = b2-4ac
Δ = 2482-4·(-8)·640
Δ = 81984
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:

$\sqrt{\Delta}=\sqrt{81984}=\sqrt{64*1281}=\sqrt{64}*\sqrt{1281}=8\sqrt{1281}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(248)-8\sqrt{1281}}{2*-8}=\frac{-248-8\sqrt{1281}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(248)+8\sqrt{1281}}{2*-8}=\frac{-248+8\sqrt{1281}}{-16}
$