Find all values of $c$ for the Mean Value Theorem's $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ I need to find all values of $c$ for the $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ such that $f'(c) = -10$. I have already found that $f'(c) = \frac{-130}{13}=-10$.
I thought I could find the values of $c$ by differentiating $f(x)$ which becomes $6x^2-12x-90$, and applying the Mean Value Theorem $f'(c) = \frac{f(b)-f(a)}{b - a}$ again. 
$f(b) = 6(-5)^2-12(-5)-90 = 150+60-90 = 120$
$f(a) = 6(8)^2-12(8)-90 = 384-186 = 198$
$f'(c) = \frac{120-198}{-5-8} = \frac{78}{13} = 6$ 
Then I took $6x^2-12x-90$ again, replaced $x$ with $c$, and set it equal to $6$ before solving for $c$.
$6c^2-12c-90 = 6  \to 6(c^2-2c-16) = 0$
$\frac{2 \pm \sqrt{2^2-4(1)(-17)}}{2(1)} = \frac{2 \pm \sqrt{74}}{2}$ or $c = \frac{2 - \sqrt{74}}{2}, \frac{2 + \sqrt{74}}{2}$
But this answer is wrong.
How can I find the values for $c$ here?

You got the wrong answer because you set $f'(c)=\frac{f'(b)-f'(a)}{b-a}$ instead of $\frac{f(b)-f(a)}{b-a}.$ 
Otherwise your approach was correct.
