The equation of a curve is $y = -2x^2 + 3x + 5$.

1. Factor out the coefficient of $x^2$ from the $x$ terms

$y = -2\left(x^2 - \frac{3}{2}x\right) + 5$

2. Complete the square for the expression inside the bracket

$y = -2\left[ \left(x - \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 \right] + 5$

3. Expand and simplify to reach the final form

$y = -2\left(x - \frac{3}{4}\right)^2 + 2\left(\frac{9}{16}\right) + 5$
$y = -2\left(x - \frac{3}{4}\right)^2 + \frac{9}{8} + \frac{40}{8}$
$y = -2\left(x - \frac{3}{4}\right)^2 + \frac{49}{8}$

1. State the non-negativity of the squared term

For any real value of $x$, $(x - h)^2 \geqslant 0$.

2. Explain the effect of the negative coefficient $a$ and conclude the maximum value

Since $a = -2$ (which is negative), $a(x - h)^2 \leqslant 0$. Therefore, $y = a(x - h)^2 + k \leqslant k$. The maximum value of $y$ is $k$, and this occurs when $a(x - h)^2 = 0$, which means $x - h = 0 \implies x = h$.