Finding k in a polynomial The equation in x 

$3x^4 - 16x^3+18x^2+k=0$

has real solutions when..... 

I need to find when this equation has four real solutions. I've tried 

$x^2(3x^2-16x+18) +k=0$

and using $b^2-4ac$

$256-216 = 40$

I don't know where to go from here....
Given $$3x^4-16x^3+18x^2+k=0\Rightarrow 3x^4-16x^3+18x^2 = -k$$

Now Let $$f(x) = 3x^4-16x^3+18x^2 = -k$$

So we will draw the graph of $f(x) = 3x^4-16x^3+18x^2$ and $f(x)=-k$ 

Now for Skeching graph of $$\displaystyle f(x) = 3x^4-16x^3+18x^2 $$

Now Using Second Derivative test.

$$f'(x) = 12x^3-48x^2+36x=12x(x^2-4x+3) = 12x(x-1)(x-3)$$

Now for Max. and Min., Put $f'(x) =0$

So we get $x=0,1,3$

Now $$f''(x) = 36x^2-96x+36 = 12(3x^2-8x+3)$$

Now at $x=0;,$ We get $f''(0) = 36&gt;0$

So $x=0$ is a point of local Minimum

Now at $x=1;,$ We get $f''(1) = -24&lt;0$

So $x=1$ is a point of local Maximum

Now at $x=3;,$ We get $f''(3)&gt;0$

So $x=3$ is a point of local Minimum

So The graph of the function is like this.



So we get $f(0) = 0$ and $f(1) = 5$

So for $4$ real solution  line $f(x)=-k$ intersect the graph of $f(x)=0$ at exactly $4$ points

So from graph $0&lt;-k&lt;5\Rightarrow -5 &lt;k&lt;0\Rightarrow k\in (-5,0)$
The answer is $k\in (-5,0)$. Critical points are comming from the square root $\sqrt{-k^3 + 22 k^2 + 135 k}=0$ which gives $-5,0,27$ as suspicious points. Then you check in between them and find the unique interval. You can also find the suspicious points from the inverse function theorem - find the values of $k(x)$ where $k'(x)=0$.