3m(4+6m)+m(m+1)=0

Solution for 3m(4+6m)+m(m+1)=0 equation:

3m(4+6m)+m(m+1)=0

We add all the numbers together, and all the variables

3m(6m+4)+m(m+1)=0

We multiply parentheses

18m^2+m^2+12m+m=0

We add all the numbers together, and all the variables

19m^2+13m=0

a = 19; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·19·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*19}=\frac{-26}{38}
=-13/19
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*19}=\frac{0}{38}
=0
$