Q.Determine the range of the expression $\dfrac{x+2}{2x^2+3x+6}$.

Set $y$ equal to the expression, clear denominators to get a quadratic in $x$, and require its discriminant to be $\ge0$ for real $x$; that inequality in $y$ gives the range.

Let $y = \dfrac{x+2}{2x^2+3x+6}$. Note the denominator $2x^2+3x+6$ has discriminant $9-48=-39<0$, so it is never zero for real $x$ — the domain is all reals.

Cross-multiplying:

$y(2x^2+3x+6) = x+2 \ \Rightarrow\ 2y\,x^2+(3y-1)x+(6y-2)=0.$

Case $y=0$: the equation becomes $-x-2=0\Rightarrow x=-2$, a valid real solution, so $y=0$ is attained.

Case $y\ne0$: this is a genuine quadratic in $x$, so real $x$ requires discriminant $\ge0$:

$(3y-1)^2-4(2y)(6y-2)\ge0$

$9y^2-6y+1-48y^2+16y\ge0$

$-39y^2+10y+1\ge0 \ \Longleftrightarrow\ 39y^2-10y-1\le0.$

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