Answer
$3$ zeros.
Work Step by Step
$(2t-1)^2-t^4=0\\4t^2-4t+1-t^4=0\\t^4-4t^2+4t-1=0\\(t^2+2t-1)(t^2-2t+1)=0\\(t^2+2t-1)(t-1)^2=0$.
Thus $t^2+2t-1=0\\(t+1)^2-1-1=0\\(t+1)^2=2\\t+1=\pm\sqrt2\\t=-1\pm\sqrt2$ or $(t-1)^2=0\\t-1=0\\t=1$
Thus the polynomial has $3$ zeros.
