1/4m+8=2m-12

Solution for 1/4m+8=2m-12 equation:

1/4m+8=2m-12

We move all terms to the left:

1/4m+8-(2m-12)=0


Domain of the equation: 4m!=0

m!=0/4

m!=0

m∈R


We get rid of parentheses

1/4m-2m+12+8=0

We multiply all the terms by the denominator


-2m*4m+12*4m+8*4m+1=0

Wy multiply elements

-8m^2+48m+32m+1=0

We add all the numbers together, and all the variables

-8m^2+80m+1=0

a = -8; b = 80; c = +1;
Δ = b2-4ac
Δ = 802-4·(-8)·1
Δ = 6432
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:

$\sqrt{\Delta}=\sqrt{6432}=\sqrt{16*402}=\sqrt{16}*\sqrt{402}=4\sqrt{402}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-4\sqrt{402}}{2*-8}=\frac{-80-4\sqrt{402}}{-16}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+4\sqrt{402}}{2*-8}=\frac{-80+4\sqrt{402}}{-16}
$