3/2x+3=9+x

Solution for 3/2x+3=9+x equation:

3/2x+3=9+x

We move all terms to the left:

3/2x+3-(9+x)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We add all the numbers together, and all the variables

3/2x-(x+9)+3=0

We get rid of parentheses

3/2x-x-9+3=0

We multiply all the terms by the denominator


-x*2x-9*2x+3*2x+3=0

Wy multiply elements

-2x^2-18x+6x+3=0

We add all the numbers together, and all the variables

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

a = -2; b = -12; c = +3;
Δ = b2-4ac
Δ = -122-4·(-2)·3
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{42}}{2*-2}=\frac{12-2\sqrt{42}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{42}}{2*-2}=\frac{12+2\sqrt{42}}{-4}
$