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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

-6m^2+8m-(2m(4m-3))=0


We calculate terms in parentheses: -(2m(4m-3)), so:

2m(4m-3)

We multiply parentheses

8m^2-6m

Back to the equation:

-(8m^2-6m)


We get rid of parentheses

-6m^2-8m^2+8m+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{-28}{-28}
=1
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*-14}=\frac{0}{-28}
=0
$