3/4m-9=3+m

Solution for 3/4m-9=3+m equation:

3/4m-9=3+m

We move all terms to the left:

3/4m-9-(3+m)=0


Domain of the equation: 4m!=0

m!=0/4

m!=0

m∈R


We add all the numbers together, and all the variables

3/4m-(m+3)-9=0

We get rid of parentheses

3/4m-m-3-9=0

We multiply all the terms by the denominator


-m*4m-3*4m-9*4m+3=0

Wy multiply elements

-4m^2-12m-36m+3=0

We add all the numbers together, and all the variables

-4m^2-48m+3=0

a = -4; b = -48; c = +3;
Δ = b2-4ac
Δ = -482-4·(-4)·3
Δ = 2352
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:

$\sqrt{\Delta}=\sqrt{2352}=\sqrt{784*3}=\sqrt{784}*\sqrt{3}=28\sqrt{3}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-28\sqrt{3}}{2*-4}=\frac{48-28\sqrt{3}}{-8}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+28\sqrt{3}}{2*-4}=\frac{48+28\sqrt{3}}{-8}
$