Number of real roots of $3x^4+6x^3+x^2+6x+3$ How many real roots does the following quartic polynomial have?
$$3x^4+6x^3+x^2+6x+3$$
   After dividing both sides by $x^2$, we get
$$3x^2+6x+1+\dfrac6x+\dfrac3{x^2}=0$$
Or,$$3\left(x^2+\dfrac1{x^2}\right)+6\left(x+\dfrac1x\right)+1=0$$
Taking $x+\dfrac1x$ as $t$
$$3t^2-2+6t+1=0$$
Or,$$3t^2+6t-1=0$$
On solving I got the roots
$$\dfrac{-3+2\sqrt6}3$$ and $$\dfrac{-3-2\sqrt6}3$$
Then I plugged in the values and found only 2 roots are real use discriminant

You have committed a mistake. The correct solution after substitution is 
$$3(t^2-2)+6t+1=0\\
\implies3t^2+6t-5=0$$
Note that it takes value $3\times4+12-5>0$ at $2$ and $-5<0$ at $-2$. Hence, there is precisely one root between $2$ and $-2$. But we have $2$ solutions of the original quartic equation for each value of $|t|\ge2$. Hence, the original quartic equation has $2$ real solutions.
