(3/2)+(3/4)x=2x+8-5x

Solution for (3/2)+(3/4)x=2x+8-5x equation:

(3/2)+(3/4)x=2x+8-5x

We move all terms to the left:

(3/2)+(3/4)x-(2x+8-5x)=0


Domain of the equation: 4)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+3/4)x-(-3x+8)+(+3/2)=0

We multiply parentheses

3x^2-(-3x+8)+(+3/2)=0

We get rid of parentheses

3x^2+3x-8+3/2=0

We multiply all the terms by the denominator


3x^2*2+3x*2+3-8*2=0

We add all the numbers together, and all the variables

3x^2*2+3x*2-13=0

Wy multiply elements

6x^2+6x-13=0

a = 6; b = 6; c = -13;
Δ = b2-4ac
Δ = 62-4·6·(-13)
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{87}}{2*6}=\frac{-6-2\sqrt{87}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{87}}{2*6}=\frac{-6+2\sqrt{87}}{12}
$