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

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

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

We move all terms to the left:

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

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

We multiply parentheses ..

-2m^2+(+8m^2+16m-6m-12)=0

We get rid of parentheses

-2m^2+8m^2+16m-6m-12=0

We add all the numbers together, and all the variables

6m^2+10m-12=0

a = 6; b = 10; c = -12;
Δ = b2-4ac
Δ = 102-4·6·(-12)
Δ = 388
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:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{97}}{2*6}=\frac{-10-2\sqrt{97}}{12}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{97}}{2*6}=\frac{-10+2\sqrt{97}}{12}
$