7/3x+5=5/64x-4

Solution for 7/3x+5=5/64x-4 equation:

7/3x+5=5/64x-4

We move all terms to the left:

7/3x+5-(5/64x-4)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 64x-4)!=0

x∈R


We get rid of parentheses

7/3x-5/64x+4+5=0

We calculate fractions


448x/192x^2+(-15x)/192x^2+4+5=0

We add all the numbers together, and all the variables

448x/192x^2+(-15x)/192x^2+9=0

We multiply all the terms by the denominator


448x+(-15x)+9*192x^2=0

Wy multiply elements

1728x^2+448x+(-15x)=0

We get rid of parentheses

1728x^2+448x-15x=0

We add all the numbers together, and all the variables

1728x^2+433x=0

a = 1728; b = 433; c = 0;
Δ = b2-4ac
Δ = 4332-4·1728·0
Δ = 187489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{187489}=433$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(433)-433}{2*1728}=\frac{-866}{3456}
=-433/1728
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(433)+433}{2*1728}=\frac{0}{3456}
=0
$