(1/2x)-10+(3/2x)=x+1

Solution for (1/2x)-10+(3/2x)=x+1 equation:

(1/2x)-10+(3/2x)=x+1

We move all terms to the left:

(1/2x)-10+(3/2x)-(x+1)=0


Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+1/2x)+(+3/2x)-(x+1)-10=0

We get rid of parentheses

1/2x+3/2x-x-1-10=0

We multiply all the terms by the denominator


-x*2x-1*2x-10*2x+1+3=0

We add all the numbers together, and all the variables

-x*2x-1*2x-10*2x+4=0

Wy multiply elements

-2x^2-2x-20x+4=0

We add all the numbers together, and all the variables

-2x^2-22x+4=0

a = -2; b = -22; c = +4;
Δ = b2-4ac
Δ = -222-4·(-2)·4
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{129}}{2*-2}=\frac{22-2\sqrt{129}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{129}}{2*-2}=\frac{22+2\sqrt{129}}{-4}
$