Question 148290
{{{2t^(-4)-5t^(-2)+2=0}}}
This is a quadratic in {{{t^(-2)}}}
Sub x for t^-2, and it's
{{{2x^2-5x-2=0}}}
*[invoke solve_quadratic_equation 2,-5,2]
x=1/2, x=2
x is t^(-2), so
{{{1/t^2=1/2}}}
{{{t^2 = 2}}}
{{{t=sqrt(2)}}},{{{t=-sqrt(2)}}}
The 2nd root is the inverse, so the other 2 solutions are the inverse of the 2 shown, or
{{{t=+sqrt(2)/2}}} and {{{t=-sqrt(2)/2}}}