5/2x-3=7/2x-x+5

Solution for 5/2x-3=7/2x-x+5 equation:

5/2x-3=7/2x-x+5

We move all terms to the left:

5/2x-3-(7/2x-x+5)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R



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

x∈R


We add all the numbers together, and all the variables

5/2x-(-1x+7/2x+5)-3=0

We get rid of parentheses

5/2x+1x-7/2x-5-3=0

We multiply all the terms by the denominator


1x*2x-5*2x-3*2x+5-7=0

We add all the numbers together, and all the variables

1x*2x-5*2x-3*2x-2=0

Wy multiply elements

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

We add all the numbers together, and all the variables

2x^2-16x-2=0

a = 2; b = -16; c = -2;
Δ = b2-4ac
Δ = -162-4·2·(-2)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{17}}{2*2}=\frac{16-4\sqrt{17}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{17}}{2*2}=\frac{16+4\sqrt{17}}{4}
$