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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

6m^2+12m+2m+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}
$