Answer
No real solution.
Work Step by Step
The given expression is
$4t^2+t+1=0$
The equation is in standard form $at^2+bt+c=0$.
We have $a=4,b=1$ and $c=1$.
The discriminant is
$=b^2-4ac$
$=(1)^2-4(4)(1)$
$=1-16$
$=-15$
Since $-15<0$. There is no real solution.
Hence, the equation has no real solution.
