3x+10=7/2x+5

Solution for 3x+10=7/2x+5 equation:

3x+10=7/2x+5

We move all terms to the left:

3x+10-(7/2x+5)=0


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

x∈R


We get rid of parentheses

3x-7/2x-5+10=0

We multiply all the terms by the denominator


3x*2x-5*2x+10*2x-7=0

Wy multiply elements

6x^2-10x+20x-7=0

We add all the numbers together, and all the variables

6x^2+10x-7=0

a = 6; b = 10; c = -7;
Δ = b2-4ac
Δ = 102-4·6·(-7)
Δ = 268
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:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{67}}{2*6}=\frac{-10-2\sqrt{67}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{67}}{2*6}=\frac{-10+2\sqrt{67}}{12}
$