First of all, is y in the denominator of 5/25y or is 5/25 just the coefficient of y?

Second of all, I don't understand how you got 25x = (y^2+5)/y, if you multiplied x = y^2+5/25y by 25, then y^2 should have a coefficient of 25 as well since it wasn't in the numerator.

It seems to me that you're trying to find the formula for y from x=y^2+1/5y.

Then a quadratic equation is enough. Just bringing the x as a constant in terms of y we get y^2+1/5y-x = 0 which gives by the quadratic formula: y = (-1/5+sqrt(1/25+4x))/2 and (-1/5-sqrt(1/25+4x))/2 as the 2 roots. That's your answer

solve for y which is the inverse

x = y^2 +5/25y

First of all, is y in the denominator of 5/25y or is 5/25 just the coefficient of y?

Second of all, I don't understand how you got 25x = (y^2+5)/y, if you multiplied x = y^2+5/25y by 25, then y^2 should have a coefficient of 25 as well since it wasn't in the numerator.

It seems to me that you're trying to find the formula for y from x=y^2+1/5y.

Then a quadratic equation is enough. Just bringing the x as a constant in terms of y we get y^2+1/5y-x = 0 which gives by the quadratic formula: y = (-1/5+sqrt(1/25+4x))/2 and (-1/5-sqrt(1/25+4x))/2 as the 2 roots. That's your answer