2(3x-4)=41/2x-8

Solution for 2(3x-4)=41/2x-8 equation:

2(3x-4)=41/2x-8

We move all terms to the left:

2(3x-4)-(41/2x-8)=0


Domain of the equation: 2x-8)!=0

x∈R


We multiply parentheses

6x-(41/2x-8)-8=0

We get rid of parentheses

6x-41/2x+8-8=0

We multiply all the terms by the denominator


6x*2x+8*2x-8*2x-41=0

Wy multiply elements

12x^2+16x-16x-41=0

We add all the numbers together, and all the variables

12x^2-41=0

a = 12; b = 0; c = -41;
Δ = b2-4ac
Δ = 02-4·12·(-41)
Δ = 1968
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:

$\sqrt{\Delta}=\sqrt{1968}=\sqrt{16*123}=\sqrt{16}*\sqrt{123}=4\sqrt{123}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{123}}{2*12}=\frac{0-4\sqrt{123}}{24}
=-\frac{4\sqrt{123}}{24}
=-\frac{\sqrt{123}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{123}}{2*12}=\frac{0+4\sqrt{123}}{24}
=\frac{4\sqrt{123}}{24}
=\frac{\sqrt{123}}{6}
$