x/2^-3=x/3^+5

Solution for x/2^-3=x/3^+5 equation:

x/2^-3=x/3^+5

We move all terms to the left:

x/2^-3-(x/3^+5)=0

We get rid of parentheses

x/2^-x/3^-5-3=0

We calculate fractions


(-2x^2)/()+3x^2/()-5-3=0

We add all the numbers together, and all the variables

(-2x^2)/()+3x^2/()-8=0

We multiply all the terms by the denominator


(-2x^2)+3x^2-8*()=0

We add all the numbers together, and all the variables

3x^2+(-2x^2)=0

We get rid of parentheses

3x^2-2x^2=0

We add all the numbers together, and all the variables

x^2=0

a = 1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·1·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{2}=0$