3/2=-7x+93x/3x

Solution for 3/2=-7x+93x/3x equation:

3/2=-7x+93x/3x

We move all terms to the left:

3/2-(-7x+93x/3x)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We get rid of parentheses

7x-93x/3x+3/2=0

We calculate fractions


7x+(-186x)/6x+9x/6x=0

We multiply all the terms by the denominator


7x*6x+(-186x)+9x=0

We add all the numbers together, and all the variables

9x+7x*6x+(-186x)=0

Wy multiply elements

42x^2+9x+(-186x)=0

We get rid of parentheses

42x^2+9x-186x=0

We add all the numbers together, and all the variables

42x^2-177x=0

a = 42; b = -177; c = 0;
Δ = b2-4ac
Δ = -1772-4·42·0
Δ = 31329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{31329}=177$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-177)-177}{2*42}=\frac{0}{84}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-177)+177}{2*42}=\frac{354}{84}
=4+3/14
$