3m^2-19m+16m=0

Solution for 3m^2-19m+16m=0 equation:

3m^2-19m+16m=0

We add all the numbers together, and all the variables

3m^2-3m=0

a = 3; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·3·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*3}=\frac{0}{6}
=0
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*3}=\frac{6}{6}
=1
$