Find the range of $f(x) = 3x^4 - 16x^3 + 18x^2 + 5$ without applying differential calculus 
Find the range of $f(x) = 3x^4 - 16x^3 + 18x^2 + 5$ without applying differential calculus.

I tried to express $$f(x)=3x^4-16x^3+18x^2+5=A(ax^2+bx+c)^2+B(ax^2+bx+c)+C $$ which is a quadratic in $ax^2+bx+c$ which itself is quadratic in $x$. Comparing coefficients, we get
$$Aa^2=3 \tag{1}$$
$$2abA=-16$$
$$A(b^2+2ac)+aB=18$$
$$2bcA+bB=0$$
$$Ac^2+Bc+C=5$$
But I felt its very lengthy to solve these equations. Any hints?

The range is $[k,+\infty)$ where $k$ is the minimum value  such that the inequality
$$
3x^4-16x^3+18x^2+5\ge k
$$
is true for any $x \in \mathbb{R}$ and this is the minimum value $k$ such that the equation 
$$
3x^4-16x^3+18x^2+5- k=0
$$
has a double solution, that is the discriminant of $3x^4-16x^3+18x^2+5- k$ is null.
The calculation of the discriminant for a quartic polynomial is a bit ''heavy'', but Wolfram Alpha gives:
$$\Delta= -6912(k-5)(k-10)(k+22)$$
so the minimum value is $k=-22$. 
