m+3m(m-4)=16

Solution for m+3m(m-4)=16 equation:

m+3m(m-4)=16

We move all terms to the left:

m+3m(m-4)-(16)=0

We multiply parentheses

3m^2+m-12m-16=0

We add all the numbers together, and all the variables

3m^2-11m-16=0

a = 3; b = -11; c = -16;
Δ = b2-4ac
Δ = -112-4·3·(-16)
Δ = 313
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{313}}{2*3}=\frac{11-\sqrt{313}}{6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{313}}{2*3}=\frac{11+\sqrt{313}}{6}
$