3w(5w+9)=0

Solution for 3w(5w+9)=0 equation:

3w(5w+9)=0

We multiply parentheses

15w^2+27w=0

a = 15; b = 27; c = 0;
Δ = b2-4ac
Δ = 272-4·15·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-27}{2*15}=\frac{-54}{30}
=-1+4/5
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+27}{2*15}=\frac{0}{30}
=0
$