2(2m-5)+2(4m2)=396

Solution for 2(2m-5)+2(4m2)=396 equation:

2(2m-5)+2(4m^2)=396

We move all terms to the left:

2(2m-5)+2(4m^2)-(396)=0

determiningTheFunctionDomain
24m^2+2(2m-5)-396=0

We multiply parentheses

24m^2+4m-10-396=0

We add all the numbers together, and all the variables

24m^2+4m-406=0

a = 24; b = 4; c = -406;
Δ = b2-4ac
Δ = 42-4·24·(-406)
Δ = 38992
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:

$\sqrt{\Delta}=\sqrt{38992}=\sqrt{16*2437}=\sqrt{16}*\sqrt{2437}=4\sqrt{2437}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{2437}}{2*24}=\frac{-4-4\sqrt{2437}}{48}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{2437}}{2*24}=\frac{-4+4\sqrt{2437}}{48}
$