160=(12+x)x

Solution for 160=(12+x)x equation:

160=(12+x)x

We move all terms to the left:

160-((12+x)x)=0

We add all the numbers together, and all the variables

-((x+12)x)+160=0


We calculate terms in parentheses: -((x+12)x), so:

(x+12)x

We multiply parentheses

x^2+12x

Back to the equation:

-(x^2+12x)


We get rid of parentheses

-x^2-12x+160=0

We add all the numbers together, and all the variables

-1x^2-12x+160=0

a = -1; b = -12; c = +160;
Δ = b2-4ac
Δ = -122-4·(-1)·160
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-28}{2*-1}=\frac{-16}{-2}
=+8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+28}{2*-1}=\frac{40}{-2}
=-20
$