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

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

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

We move all terms to the left:

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


Domain of the equation: 2x+3)!=0

x∈R



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

x∈R


We get rid of parentheses

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

We calculate fractions


3x/6x^2+4x/6x^2+3-5-180=0

We add all the numbers together, and all the variables

3x/6x^2+4x/6x^2-182=0

We multiply all the terms by the denominator


3x+4x-182*6x^2=0

We add all the numbers together, and all the variables

7x-182*6x^2=0

Wy multiply elements

-1092x^2+7x=0

a = -1092; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·(-1092)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*-1092}=\frac{-14}{-2184}
=1/156
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*-1092}=\frac{0}{-2184}
=0
$