Question 945371
This is a quadratic equation.
{{{z=x^(5/8)}}}

Then
{{{z^2=x^(5/4)}}}
So,
{{{z^2-14z+49=0}}}
{{{(z-7)^2=0}}}
{{{z-7=0}}}
{{{z=7}}}
Substituting back,
{{{highlight(x=7^(8/5))}}}
That is, {{{7}}} to the {{{8/5}}} power.