17/3x=4/12x+48

Solution for 17/3x=4/12x+48 equation:

17/3x=4/12x+48

We move all terms to the left:

17/3x-(4/12x+48)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 12x+48)!=0

x∈R


We get rid of parentheses

17/3x-4/12x-48=0

We calculate fractions


204x/36x^2+(-12x)/36x^2-48=0

We multiply all the terms by the denominator


204x+(-12x)-48*36x^2=0

Wy multiply elements

-1728x^2+204x+(-12x)=0

We get rid of parentheses

-1728x^2+204x-12x=0

We add all the numbers together, and all the variables

-1728x^2+192x=0

a = -1728; b = 192; c = 0;
Δ = b2-4ac
Δ = 1922-4·(-1728)·0
Δ = 36864
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36864}=192$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(192)-192}{2*-1728}=\frac{-384}{-3456}
=1/9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(192)+192}{2*-1728}=\frac{0}{-3456}
=0
$