Question 76303
First get all terms to one side

{{{25y^2-20y+4=0}}}

Now factor the left side

*[invoke quadratic_factoring 25, -20, 4]

If we replace the x's with y's, the left side then becomes

{{{(5y-2)(5y-2)=0}}}
Since the two factors are the same, we can write them as this
{{{(5y-2)^2=0}}} 
{{{sqrt(5x-2)^2=sqrt(0)}}}  Take the square root of both sides
{{{5y-2=0}}} 
{{{5y=2}}} Add 2 to both sides
{{{y=2/5}}} Divide both sides by 5

So our answer is {{{y=2/5}}}

<p>
Check:
{{{25y^2-20y+4=0}}} 
{{{25(2/5)^2-20(2/5)+4=0}}} Plug in {{{y=2/5}}}
{{{25(4/25)-40/5+4=0}}}
{{{100/25-40/5+4=0}}}
{{{4-8+4=0}}}
{{{8-8=0}}}
{{{0=0}}} works