3r(2r+9)=0

Solution for 3r(2r+9)=0 equation:

3r(2r+9)=0

We multiply parentheses

6r^2+27r=0

a = 6; b = 27; c = 0;
Δ = b2-4ac
Δ = 272-4·6·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-27}{2*6}=\frac{-54}{12}
=-4+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+27}{2*6}=\frac{0}{12}
=0
$