3g(2g+5)=0

Solution for 3g(2g+5)=0 equation:

3g(2g+5)=0

We multiply parentheses

6g^2+15g=0

a = 6; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·6·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*6}=\frac{-30}{12}
=-2+1/2
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*6}=\frac{0}{12}
=0
$