56=r^2+10r

Solution for 56=r^2+10r equation:

56=r^2+10r

We move all terms to the left:

56-(r^2+10r)=0

We get rid of parentheses

-r^2-10r+56=0

We add all the numbers together, and all the variables

-1r^2-10r+56=0

a = -1; b = -10; c = +56;
Δ = b2-4ac
Δ = -102-4·(-1)·56
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-18}{2*-1}=\frac{-8}{-2}
=+4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+18}{2*-1}=\frac{28}{-2}
=-14
$