7/3x+7x+7-x+3=17+4x-13

Solution for 7/3x+7x+7-x+3=17+4x-13 equation:

7/3x+7x+7-x+3=17+4x-13

We move all terms to the left:

7/3x+7x+7-x+3-(17+4x-13)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

7/3x+7x-x-(4x+4)+7+3=0

We add all the numbers together, and all the variables

6x+7/3x-(4x+4)+10=0

We get rid of parentheses

6x+7/3x-4x-4+10=0

We multiply all the terms by the denominator


6x*3x-4x*3x-4*3x+10*3x+7=0

Wy multiply elements

18x^2-12x^2-12x+30x+7=0

We add all the numbers together, and all the variables

6x^2+18x+7=0

a = 6; b = 18; c = +7;
Δ = b2-4ac
Δ = 182-4·6·7
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{39}}{2*6}=\frac{-18-2\sqrt{39}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{39}}{2*6}=\frac{-18+2\sqrt{39}}{12}
$