2/7x+3=9/2x-5

Solution for 2/7x+3=9/2x-5 equation:

2/7x+3=9/2x-5

We move all terms to the left:

2/7x+3-(9/2x-5)=0


Domain of the equation: 7x!=0

x!=0/7

x!=0

x∈R



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

x∈R


We get rid of parentheses

2/7x-9/2x+5+3=0

We calculate fractions


4x/14x^2+(-63x)/14x^2+5+3=0

We add all the numbers together, and all the variables

4x/14x^2+(-63x)/14x^2+8=0

We multiply all the terms by the denominator


4x+(-63x)+8*14x^2=0

Wy multiply elements

112x^2+4x+(-63x)=0

We get rid of parentheses

112x^2+4x-63x=0

We add all the numbers together, and all the variables

112x^2-59x=0

a = 112; b = -59; c = 0;
Δ = b2-4ac
Δ = -592-4·112·0
Δ = 3481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3481}=59$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-59)-59}{2*112}=\frac{0}{224}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-59)+59}{2*112}=\frac{118}{224}
=59/112
$