(3x-20)+(2/3x)=90

Solution for (3x-20)+(2/3x)=90 equation:

(3x-20)+(2/3x)=90

We move all terms to the left:

(3x-20)+(2/3x)-(90)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(3x-20)+(+2/3x)-90=0

We get rid of parentheses

3x+2/3x-20-90=0

We multiply all the terms by the denominator


3x*3x-20*3x-90*3x+2=0

Wy multiply elements

9x^2-60x-270x+2=0

We add all the numbers together, and all the variables

9x^2-330x+2=0

a = 9; b = -330; c = +2;
Δ = b2-4ac
Δ = -3302-4·9·2
Δ = 108828
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:

$\sqrt{\Delta}=\sqrt{108828}=\sqrt{36*3023}=\sqrt{36}*\sqrt{3023}=6\sqrt{3023}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-330)-6\sqrt{3023}}{2*9}=\frac{330-6\sqrt{3023}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-330)+6\sqrt{3023}}{2*9}=\frac{330+6\sqrt{3023}}{18}
$