The equation of a curve is $y = 3x^2 + 5x + 1$. Find the set of values of $x$ for which the curve lies completely above the line $y - 3x = 2$. Find the range of values of $m$ for which the equation $mx^2 + 2m = 3x(4 - x)$ has real roots.

1. Set up the inequality for the curve lying above the line

The curve lies above the line when $y_{\text{curve}} > y_{\text{line}}$.
$3x^2 + 5x + 1 > 3x + 2$

2. Rearrange into a standard quadratic inequality and factorise

$3x^2 + 2x - 1 > 0$
$(3x - 1)(x + 1) > 0$

3. Solve the quadratic inequality

The critical values are $x = \frac{1}{3}$ and $x = -1$.
Since the quadratic is positive, the solution is:
$x < -1 \text{ or } x > \frac{1}{3}$

1. Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$.

$mx^2 + 2m = 12x - 3x^2$
$(m + 3)x^2 - 12x + 2m = 0$

2. Apply the discriminant condition for real roots.

For real roots, $b^2 - 4ac \geqslant 0$.
$(-12)^2 - 4(m + 3)(2m) \geqslant 0$

3. Expand and simplify the inequality.

$144 - 8m(m + 3) \geqslant 0$
$144 - 8m^2 - 24m \geqslant 0$
$8m^2 + 24m - 144 \leqslant 0$
$m^2 + 3m - 18 \leqslant 0$

4. Solve the quadratic inequality for $m$.

$(m + 6)(m - 3) \leqslant 0$
$-6 \leqslant m \leqslant 3$