3/4x-9=3+x

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

3/4x-9=3+x

We move all terms to the left:

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


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

3/4x-x-3-9=0

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

-4x^2-48x+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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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