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

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

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

We move all terms to the left:

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


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

x∈R



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

x∈R


We get rid of parentheses

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

We calculate fractions


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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

5x-4*2x^2=0

Wy multiply elements

-8x^2+5x=0

a = -8; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-8)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-8}=\frac{-10}{-16}
=5/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-8}=\frac{0}{-16}
=0
$