3x+20=7/4x+8

Solution for 3x+20=7/4x+8 equation:

3x+20=7/4x+8

We move all terms to the left:

3x+20-(7/4x+8)=0


Domain of the equation: 4x+8)!=0

x∈R


We get rid of parentheses

3x-7/4x-8+20=0

We multiply all the terms by the denominator


3x*4x-8*4x+20*4x-7=0

Wy multiply elements

12x^2-32x+80x-7=0

We add all the numbers together, and all the variables

12x^2+48x-7=0

a = 12; b = 48; c = -7;
Δ = b2-4ac
Δ = 482-4·12·(-7)
Δ = 2640
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:

$\sqrt{\Delta}=\sqrt{2640}=\sqrt{16*165}=\sqrt{16}*\sqrt{165}=4\sqrt{165}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-4\sqrt{165}}{2*12}=\frac{-48-4\sqrt{165}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+4\sqrt{165}}{2*12}=\frac{-48+4\sqrt{165}}{24}
$