r(r+5)=2

Solution for r(r+5)=2 equation:

r(r+5)=2

We move all terms to the left:

r(r+5)-(2)=0

We multiply parentheses

r^2+5r-2=0

a = 1; b = 5; c = -2;
Δ = b2-4ac
Δ = 52-4·1·(-2)
Δ = 33
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{-(5)-\sqrt{33}}{2*1}=\frac{-5-\sqrt{33}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{33}}{2*1}=\frac{-5+\sqrt{33}}{2}
$