Question 175737
Your first problem is that you don't actually have an equation.  There is not an equal sign to be seen anywhere in {{{y^4-14y^2+45}}}.  So all I can do is assume that you meant {{{y^4 - 14y^2 + 45 = 0}}}.  Presuming this is the case, let {{{t = y^2}}}


Then you can write:  {{{t^2 - 14t + 45 = 0}}} which is a quadratic that can be solved by ordinary means; this one factors because {{{ (-5) * (-9) = 45}}} and {{{(-5) + (-9) = -14}}}, so:


{{{t^2 - 14t + 45 = (t - 5)(t - 9) = 0}}}


Therefore, {{{t - 5 = 0}}} or {{{t - 9 = 0}}} which is to say:


{{{t = 5}}} or {{{t = 9}}}


But remember that {{{t = y^2}}}, so we can now say:


{{{y^2 = 5}}} which means that {{{y = sqrt(5)}}} or {{{y = -sqrt(5)}}}, and


{{{y^2 = 9}}} which means that {{{y = sqrt(9) = 3}}} or {{{y = -sqrt(9) = -3}}}


And there are your four real number roots.