3x+35=7/2x-5

Solution for 3x+35=7/2x-5 equation:

3x+35=7/2x-5

We move all terms to the left:

3x+35-(7/2x-5)=0


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

x∈R


We get rid of parentheses

3x-7/2x+5+35=0

We multiply all the terms by the denominator


3x*2x+5*2x+35*2x-7=0

Wy multiply elements

6x^2+10x+70x-7=0

We add all the numbers together, and all the variables

6x^2+80x-7=0

a = 6; b = 80; c = -7;
Δ = b2-4ac
Δ = 802-4·6·(-7)
Δ = 6568
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:

$\sqrt{\Delta}=\sqrt{6568}=\sqrt{4*1642}=\sqrt{4}*\sqrt{1642}=2\sqrt{1642}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-2\sqrt{1642}}{2*6}=\frac{-80-2\sqrt{1642}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+2\sqrt{1642}}{2*6}=\frac{-80+2\sqrt{1642}}{12}
$