10=r^2+r

Solution for 10=r^2+r equation:

10=r^2+r

We move all terms to the left:

10-(r^2+r)=0

We get rid of parentheses

-r^2-r+10=0

We add all the numbers together, and all the variables

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

a = -1; b = -1; c = +10;
Δ = b2-4ac
Δ = -12-4·(-1)·10
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{41}}{2*-1}=\frac{1-\sqrt{41}}{-2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{41}}{2*-1}=\frac{1+\sqrt{41}}{-2}
$