3/2x+x+(x-2)=40

Solution for 3/2x+x+(x-2)=40 equation:

3/2x+x+(x-2)=40

We move all terms to the left:

3/2x+x+(x-2)-(40)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We add all the numbers together, and all the variables

x+3/2x+(x-2)-40=0

We get rid of parentheses

x+3/2x+x-2-40=0

We multiply all the terms by the denominator


x*2x+x*2x-2*2x-40*2x+3=0

Wy multiply elements

2x^2+2x^2-4x-80x+3=0

We add all the numbers together, and all the variables

4x^2-84x+3=0

a = 4; b = -84; c = +3;
Δ = b2-4ac
Δ = -842-4·4·3
Δ = 7008
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:

$\sqrt{\Delta}=\sqrt{7008}=\sqrt{16*438}=\sqrt{16}*\sqrt{438}=4\sqrt{438}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-84)-4\sqrt{438}}{2*4}=\frac{84-4\sqrt{438}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-84)+4\sqrt{438}}{2*4}=\frac{84+4\sqrt{438}}{8}
$