(4/3x)+(12x-20)=180

Solution for (4/3x)+(12x-20)=180 equation:

(4/3x)+(12x-20)=180

We move all terms to the left:

(4/3x)+(12x-20)-(180)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+4/3x)+(12x-20)-180=0

We get rid of parentheses

4/3x+12x-20-180=0

We multiply all the terms by the denominator


12x*3x-20*3x-180*3x+4=0

Wy multiply elements

36x^2-60x-540x+4=0

We add all the numbers together, and all the variables

36x^2-600x+4=0

a = 36; b = -600; c = +4;
Δ = b2-4ac
Δ = -6002-4·36·4
Δ = 359424
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:

$\sqrt{\Delta}=\sqrt{359424}=\sqrt{9216*39}=\sqrt{9216}*\sqrt{39}=96\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-600)-96\sqrt{39}}{2*36}=\frac{600-96\sqrt{39}}{72}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-600)+96\sqrt{39}}{2*36}=\frac{600+96\sqrt{39}}{72}
$