Question 480707
Let {{{u=x^2}}} So {{{(u^2)^2=x^4}}}

{{{15x^4 - 28x^2 + 5=0}}}
{{{15u^2-28u+5=0}}}


Using the Quadratic Formula:
{{{a=15}}} {{{b=-28}}} {{{c=5}}}
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
So
{{{x= (-(-28)+-sqrt((-28)^2-4*15*5))/(2*15)}}}
{{{x=(28+sqrt(784-300))/30}}} and {{{x=(28-sqrt(784-300))/30}}} 
{{{x=(28+sqrt(484))/30}}} and {{{x=(28-sqrt(484))/30}}} 
{{{x=(28+22)/30}}} and {{{x=(28-22)/30}}}
{{{x=50/30}}} and {{{x=6/30}}}
{{{x=5/3}}} and {{{x=1/5}}}

remember {{{u = x^2}}} so:

{{{x = sqrt(5/3) = sqrt(15)/3}}} or {{{x = -sqrt(5/3) = -sqrt(15)/3}}}

or 

{{{x = sqrt(1/5) = sqrt(5)/5}}} or {{{x = -sqrt(1/5) = -sqrt(5)/5}}}