The equation of a curve is $y = 2x^2 - 6x + 9$.

1. Express the quadratic expression in the form $a(x+b)^2 + c$ by completing the square

$2x^2 - 6x + 9$
$= 2\left(x^2 - 3x\right) + 9$
$= 2\left[\left(x - \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2\right] + 9$
$= 2\left(x - \frac{3}{2}\right)^2 - 2\left(\frac{9}{4}\right) + 9$
$= 2\left(x - \frac{3}{2}\right)^2 - \frac{9}{2} + 9$
$= 2\left(x - \frac{3}{2}\right)^2 + \frac{9}{2}$

2. Explain whether the curve can have a value smaller than $9/2$

Since $\left(x - \frac{3}{2}\right)^2 \geqslant 0$ for all real values of $x$,
$2\left(x - \frac{3}{2}\right)^2 \geqslant 0$
$2\left(x - \frac{3}{2}\right)^2 + \frac{9}{2} \geqslant \frac{9}{2}$
The minimum value of the curve is $\frac{9}{2}$. Therefore, it is not possible for the curve to have a value smaller than $\frac{9}{2}$.

1. Set up the equation to find the points of intersection

Equate the line and the curve:
$2x^2 - 6x + 9 = 3x + 5$

2. Rearrange into a standard quadratic equation

$2x^2 - 9x + 4 = 0$

3. Solve the quadratic equation for x

$(2x - 1)(x - 4) = 0$
$x = \frac{1}{2} \text{ or } x = 4$

4. Find the corresponding y-coordinates

When $x = \frac{1}{2}$, $y = 3(\frac{1}{2}) + 5 = \frac{13}{2}$
When $x = 4$, $y = 3(4) + 5 = 17$
The points are $P(\frac{1}{2}, \frac{13}{2})$ and $Q(4, 17)$.

5. Calculate the exact length of PQ

$PQ = \sqrt{(4 - \frac{1}{2})^2 + (17 - \frac{13}{2})^2}$
$PQ = \sqrt{(\frac{7}{2})^2 + (\frac{21}{2})^2}$
$PQ = \sqrt{\frac{49}{4} + \frac{441}{4}}$
$PQ = \sqrt{\frac{490}{4}}$
$PQ = \frac{\sqrt{490}}{2} = \frac{7\sqrt{10}}{2}$