5m(3m+4)=0

Solution for 5m(3m+4)=0 equation:

5m(3m+4)=0

We multiply parentheses

15m^2+20m=0

a = 15; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·15·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*15}=\frac{-40}{30}
=-1+1/3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*15}=\frac{0}{30}
=0
$