3(2x-5)+4=2/7x+8

Solution for 3(2x-5)+4=2/7x+8 equation:

3(2x-5)+4=2/7x+8

We move all terms to the left:

3(2x-5)+4-(2/7x+8)=0


Domain of the equation: 7x+8)!=0

x∈R


We multiply parentheses

6x-(2/7x+8)-15+4=0

We get rid of parentheses

6x-2/7x-8-15+4=0

We multiply all the terms by the denominator


6x*7x-8*7x-15*7x+4*7x-2=0

Wy multiply elements

42x^2-56x-105x+28x-2=0

We add all the numbers together, and all the variables

42x^2-133x-2=0

a = 42; b = -133; c = -2;
Δ = b2-4ac
Δ = -1332-4·42·(-2)
Δ = 18025
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:

$\sqrt{\Delta}=\sqrt{18025}=\sqrt{25*721}=\sqrt{25}*\sqrt{721}=5\sqrt{721}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-133)-5\sqrt{721}}{2*42}=\frac{133-5\sqrt{721}}{84}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-133)+5\sqrt{721}}{2*42}=\frac{133+5\sqrt{721}}{84}
$