2m-3(m+1)=4(m+3)m+

Solution for 2m-3(m+1)=4(m+3)m+ equation:

2m-3(m+1)=4(m+3)m+

We move all terms to the left:

2m-3(m+1)-(4(m+3)m+)=0

We multiply parentheses

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


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

4(m+3)m+

We add all the numbers together, and all the variables

4(m+3)m

We multiply parentheses

4m^2+12m

Back to the equation:

-(4m^2+12m)


We add all the numbers together, and all the variables

-1m-(4m^2+12m)-3=0

We get rid of parentheses

-4m^2-1m-12m-3=0

We add all the numbers together, and all the variables

-4m^2-13m-3=0

a = -4; b = -13; c = -3;
Δ = b2-4ac
Δ = -132-4·(-4)·(-3)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-11}{2*-4}=\frac{2}{-8}
=-1/4
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+11}{2*-4}=\frac{24}{-8}
=-3
$