Question 365774
{{{10x^4-27x^2 +5=(5x^2-1)(2x^2-5) = 0}}}
Thus {{{x^2 = 1/5}}} and {{{x^2 = 5/2}}}, or 
{{{x = 1/sqrt(5)}}}, {{{x = -1/sqrt(5)}}}, {{{x = sqrt(5/2)}}}, or {{{x =- sqrt(5/2)}}}.