4(m+3)-2m=3m(m-3)

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

4(m+3)-2m=3m(m-3)

We move all terms to the left:

4(m+3)-2m-(3m(m-3))=0

We add all the numbers together, and all the variables

-2m+4(m+3)-(3m(m-3))=0

We multiply parentheses

-2m+4m-(3m(m-3))+12=0


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

3m(m-3)

We multiply parentheses

3m^2-9m

Back to the equation:

-(3m^2-9m)


We add all the numbers together, and all the variables

2m-(3m^2-9m)+12=0

We get rid of parentheses

-3m^2+2m+9m+12=0

We add all the numbers together, and all the variables

-3m^2+11m+12=0

a = -3; b = 11; c = +12;
Δ = b2-4ac
Δ = 112-4·(-3)·12
Δ = 265
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{265}}{2*-3}=\frac{-11-\sqrt{265}}{-6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{265}}{2*-3}=\frac{-11+\sqrt{265}}{-6}
$