Calculation of all values of $a$ for which $3x^2+(4-2a)x-8-a^2\leq 0$. Calculation of  all values of $a$ for which $3x^2+(4-2a)x-8-a^2\leq 0$.
Given that $x$ lies between $-3$ and $2$.
My Try:: Let $x = \alpha,\beta$ be the Roots of Given equation. where  $-3<\alpha,\beta<2$
So $\displaystyle \alpha+\beta = \frac{2a-4}{3}$ and $\displaystyle \alpha.\beta = \frac{-(8+a^2)}{3}$
Now I Did not Understand How can I proceed
Help Required
Thanks

Hint : What you have done is correct. But, don't you think it would be simple to solve the quadratic equation; this will give you the two roots of the equations as a function of parameter "a". Then apply the conditions. Are you able to continue with this ?
