x+(3/2x)+(3/2x-5)=75

Solution for x+(3/2x)+(3/2x-5)=75 equation:

x+(3/2x)+(3/2x-5)=75

We move all terms to the left:

x+(3/2x)+(3/2x-5)-(75)=0


Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R



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

x∈R


We add all the numbers together, and all the variables

x+(+3/2x)+(3/2x-5)-75=0

We get rid of parentheses

x+3/2x+3/2x-5-75=0

We multiply all the terms by the denominator


x*2x-5*2x-75*2x+3+3=0

We add all the numbers together, and all the variables

x*2x-5*2x-75*2x+6=0

Wy multiply elements

2x^2-10x-150x+6=0

We add all the numbers together, and all the variables

2x^2-160x+6=0

a = 2; b = -160; c = +6;
Δ = b2-4ac
Δ = -1602-4·2·6
Δ = 25552
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:

$\sqrt{\Delta}=\sqrt{25552}=\sqrt{16*1597}=\sqrt{16}*\sqrt{1597}=4\sqrt{1597}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-4\sqrt{1597}}{2*2}=\frac{160-4\sqrt{1597}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+4\sqrt{1597}}{2*2}=\frac{160+4\sqrt{1597}}{4}
$