3/4x+6=3+x

Solution for 3/4x+6=3+x equation:

3/4x+6=3+x

We move all terms to the left:

3/4x+6-(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)+6=0

We get rid of parentheses

3/4x-x-3+6=0

We multiply all the terms by the denominator


-x*4x-3*4x+6*4x+3=0

Wy multiply elements

-4x^2-12x+24x+3=0

We add all the numbers together, and all the variables

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

a = -4; b = 12; c = +3;
Δ = b2-4ac
Δ = 122-4·(-4)·3
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{3}}{2*-4}=\frac{-12-8\sqrt{3}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{3}}{2*-4}=\frac{-12+8\sqrt{3}}{-8}
$