60=r(10+r)

Solution for 60=r(10+r) equation:

60=r(10+r)

We move all terms to the left:

60-(r(10+r))=0

We add all the numbers together, and all the variables

-(r(r+10))+60=0


We calculate terms in parentheses: -(r(r+10)), so:

r(r+10)

We multiply parentheses

r^2+10r

Back to the equation:

-(r^2+10r)


We get rid of parentheses

-r^2-10r+60=0

We add all the numbers together, and all the variables

-1r^2-10r+60=0

a = -1; b = -10; c = +60;
Δ = b2-4ac
Δ = -102-4·(-1)·60
Δ = 340
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{85}}{2*-1}=\frac{10-2\sqrt{85}}{-2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{85}}{2*-1}=\frac{10+2\sqrt{85}}{-2}
$