r(10-r)=10

Solution for r(10-r)=10 equation:

r(10-r)=10

We move all terms to the left:

r(10-r)-(10)=0

We add all the numbers together, and all the variables

r(-1r+10)-10=0

We multiply parentheses

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

a = -1; b = 10; c = -10;
Δ = b2-4ac
Δ = 102-4·(-1)·(-10)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{15}}{2*-1}=\frac{-10-2\sqrt{15}}{-2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{15}}{2*-1}=\frac{-10+2\sqrt{15}}{-2}
$