r(r-3)=10

Solution for r(r-3)=10 equation:

r(r-3)=10

We move all terms to the left:

r(r-3)-(10)=0

We multiply parentheses

r^2-3r-10=0

a = 1; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·1·(-10)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*1}=\frac{-4}{2}
=-2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*1}=\frac{10}{2}
=5
$