r(r-3)=5

Solution for r(r-3)=5 equation:

r(r-3)=5

We move all terms to the left:

r(r-3)-(5)=0

We multiply parentheses

r^2-3r-5=0

a = 1; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·1·(-5)
Δ = 29
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{-(-3)-\sqrt{29}}{2*1}=\frac{3-\sqrt{29}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{29}}{2*1}=\frac{3+\sqrt{29}}{2}
$