2m(4m-3)=3m-7m^2+5m+m^2

Solution for 2m(4m-3)=3m-7m^2+5m+m^2 equation:

2m(4m-3)=3m-7m^2+5m+m^2

We move all terms to the left:

2m(4m-3)-(3m-7m^2+5m+m^2)=0

We multiply parentheses

-(3m-7m^2+5m+m^2)+8m^2-6m=0

We get rid of parentheses

7m^2-m^2+8m^2-3m-5m-6m=0

We add all the numbers together, and all the variables

14m^2-14m=0

a = 14; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·14·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14}{2*14}=\frac{0}{28}
=0
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*14}=\frac{28}{28}
=1
$