2(3x-4)=3/2x+1

Solution for 2(3x-4)=3/2x+1 equation:

2(3x-4)=3/2x+1

We move all terms to the left:

2(3x-4)-(3/2x+1)=0


Domain of the equation: 2x+1)!=0

x∈R


We multiply parentheses

6x-(3/2x+1)-8=0

We get rid of parentheses

6x-3/2x-1-8=0

We multiply all the terms by the denominator


6x*2x-1*2x-8*2x-3=0

Wy multiply elements

12x^2-2x-16x-3=0

We add all the numbers together, and all the variables

12x^2-18x-3=0

a = 12; b = -18; c = -3;
Δ = b2-4ac
Δ = -182-4·12·(-3)
Δ = 468
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:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{13}}{2*12}=\frac{18-6\sqrt{13}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{13}}{2*12}=\frac{18+6\sqrt{13}}{24}
$