r(2r+9)=0

Solution for r(2r+9)=0 equation:

r(2r+9)=0

We multiply parentheses

2r^2+9r=0

a = 2; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·2·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*2}=\frac{-18}{4}
=-4+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*2}=\frac{0}{4}
=0
$