(3x+17)+(1/3x-5)=180

Solution for (3x+17)+(1/3x-5)=180 equation:

(3x+17)+(1/3x-5)=180

We move all terms to the left:

(3x+17)+(1/3x-5)-(180)=0


Domain of the equation: 3x-5)!=0

x∈R


We get rid of parentheses

3x+1/3x+17-5-180=0

We multiply all the terms by the denominator


3x*3x+17*3x-5*3x-180*3x+1=0

Wy multiply elements

9x^2+51x-15x-540x+1=0

We add all the numbers together, and all the variables

9x^2-504x+1=0

a = 9; b = -504; c = +1;
Δ = b2-4ac
Δ = -5042-4·9·1
Δ = 253980
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:

$\sqrt{\Delta}=\sqrt{253980}=\sqrt{36*7055}=\sqrt{36}*\sqrt{7055}=6\sqrt{7055}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-504)-6\sqrt{7055}}{2*9}=\frac{504-6\sqrt{7055}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-504)+6\sqrt{7055}}{2*9}=\frac{504+6\sqrt{7055}}{18}
$