(3x-17)+(1/2x-5)=180

Solution for (3x-17)+(1/2x-5)=180 equation:

(3x-17)+(1/2x-5)=180

We move all terms to the left:

(3x-17)+(1/2x-5)-(180)=0


Domain of the equation: 2x-5)!=0

x∈R


We get rid of parentheses

3x+1/2x-17-5-180=0

We multiply all the terms by the denominator


3x*2x-17*2x-5*2x-180*2x+1=0

Wy multiply elements

6x^2-34x-10x-360x+1=0

We add all the numbers together, and all the variables

6x^2-404x+1=0

a = 6; b = -404; c = +1;
Δ = b2-4ac
Δ = -4042-4·6·1
Δ = 163192
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:

$\sqrt{\Delta}=\sqrt{163192}=\sqrt{4*40798}=\sqrt{4}*\sqrt{40798}=2\sqrt{40798}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-404)-2\sqrt{40798}}{2*6}=\frac{404-2\sqrt{40798}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-404)+2\sqrt{40798}}{2*6}=\frac{404+2\sqrt{40798}}{12}
$