180=(3m+1)(2m+4)

Solution for 180=(3m+1)(2m+4) equation:

180=(3m+1)(2m+4)

We move all terms to the left:

180-((3m+1)(2m+4))=0

We multiply parentheses ..

-((+6m^2+12m+2m+4))+180=0


We calculate terms in parentheses: -((+6m^2+12m+2m+4)), so:

(+6m^2+12m+2m+4)

We get rid of parentheses

6m^2+12m+2m+4

We add all the numbers together, and all the variables

6m^2+14m+4

Back to the equation:

-(6m^2+14m+4)


We get rid of parentheses

-6m^2-14m-4+180=0

We add all the numbers together, and all the variables

-6m^2-14m+176=0

a = -6; b = -14; c = +176;
Δ = b2-4ac
Δ = -142-4·(-6)·176
Δ = 4420
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:

$\sqrt{\Delta}=\sqrt{4420}=\sqrt{4*1105}=\sqrt{4}*\sqrt{1105}=2\sqrt{1105}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{1105}}{2*-6}=\frac{14-2\sqrt{1105}}{-12}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{1105}}{2*-6}=\frac{14+2\sqrt{1105}}{-12}
$