Having a bit of trouble with this one. Can anyone help?

Many thanks.The roots of the equation $\displaystyle 2x^2+6x+3=0$ are $\displaystyle \alpha$ & $\displaystyle \beta$. Show that $\displaystyle \alpha^2+\beta^2=6$.

Q.: (i)(ii)The roots of $\displaystyle 2x^2+px+q=0$ are $\displaystyle 2\alpha+\beta$ & $\displaystyle \alpha+2\beta$. Find the value of p & the value of q.

Attempt: (i)a = 2, b = 6, c = 3

$\displaystyle \alpha+\beta=\frac{-b}{a}$ => $\displaystyle \frac{-6}{2}$ => -3

$\displaystyle \alpha\beta=\frac{c}{a}$ => $\displaystyle \frac{3}{2}$

Sum of roots: $\displaystyle \alpha^2+\beta^2$ => $\displaystyle (\alpha+\beta)^2-2\alpha\beta$ => $\displaystyle (-3)^2-2(\frac{3}{2})$ => $\displaystyle 9-\frac{6}{2}$ => 9 - 3 => 6

(ii)a = 2, b = p, c = q

Sum of roots: $\displaystyle 2\alpha+\beta+\alpha+2\beta$ => $\displaystyle 3\alpha+3\beta$ => $\displaystyle 3(\alpha+\beta)$ => $\displaystyle \frac{-b}{a}$ => $\displaystyle \frac{-p}{2}$ => $\displaystyle 3(\alpha+\beta)$ => $\displaystyle 3(\frac{-p}{2})$ => $\displaystyle \frac{-3p}{2}$

Product of roots: $\displaystyle (2\alpha+\beta)(\alpha+2\beta)$ => $\displaystyle 2\alpha^2+5\alpha\beta+2\beta^2$ => $\displaystyle 2(\alpha^2+\beta^2)+5\alpha\beta$ => $\displaystyle 2((\alpha+\beta)^2-2\alpha\beta)+5\alpha\beta$ => $\displaystyle 2(\alpha+\beta)^2+\alpha\beta$ => $\displaystyle 2(\frac{-p}{2})^2+(\frac{c}{a})$ => $\displaystyle \frac{2p^2}{4}+\frac{q}{2}$ => $\displaystyle \frac{p^2}{2}+\frac{q}{2}$ => $\displaystyle \frac{p^2+q}{2}$

$\displaystyle x^2-$(sum of roots)x + (product of roots) => $\displaystyle x^2-(\frac{-3p}{2})x+\frac{p^2+q}{2}$ => $\displaystyle 2x^2+3px+p^2+q$...

Ans.:(From text book): p = 18, q = 39