r(r+3)=35

Solution for r(r+3)=35 equation:

r(r+3)=35

We move all terms to the left:

r(r+3)-(35)=0

We multiply parentheses

r^2+3r-35=0

a = 1; b = 3; c = -35;
Δ = b2-4ac
Δ = 32-4·1·(-35)
Δ = 149
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{149}}{2*1}=\frac{-3-\sqrt{149}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{149}}{2*1}=\frac{-3+\sqrt{149}}{2}
$