160=r(12+r)

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

160=r(12+r)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


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

r(r+12)

We multiply parentheses

r^2+12r

Back to the equation:

-(r^2+12r)


We get rid of parentheses

-r^2-12r+160=0

We add all the numbers together, and all the variables

-1r^2-12r+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:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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