3m(7m+12)=2(m-3)

Solution for 3m(7m+12)=2(m-3) equation:

3m(7m+12)=2(m-3)

We move all terms to the left:

3m(7m+12)-(2(m-3))=0

We multiply parentheses

21m^2+36m-(2(m-3))=0


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

2(m-3)

We multiply parentheses

2m-6

Back to the equation:

-(2m-6)


We get rid of parentheses

21m^2+36m-2m+6=0

We add all the numbers together, and all the variables

21m^2+34m+6=0

a = 21; b = 34; c = +6;
Δ = b2-4ac
Δ = 342-4·21·6
Δ = 652
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{652}=\sqrt{4*163}=\sqrt{4}*\sqrt{163}=2\sqrt{163}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-2\sqrt{163}}{2*21}=\frac{-34-2\sqrt{163}}{42}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+2\sqrt{163}}{2*21}=\frac{-34+2\sqrt{163}}{42}
$