Question 177410
{{{15x^4-28x^2+5=0}}}


The trick is to let {{{t = x^2}}}.  That gives you {{{15t^2 - 28t + 5 = 0}}} which is a quadratic in standard form. 


Look at the discriminant {{{b^2-4ac = 784 - 300 = 484 = 22^2}}}.  Since the discriminant is a perfect square, the quadratic has rational roots and can be factored:


{{{(5x - 1)(3x - 5) = 0}}}


Hence:  {{{t = 1/5}}} or {{{t = 5/3}}}


But remember {{{t = x^2}}} so:


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


or


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


Which is 4 roots as you would expect with a quartic.