已知曲线方程为 $y = 2x^2 - 6x + c$，其中 $c$ 为常数。
求使得直线 $y + 2x = 8$ 为该曲线切线的 $c$ 的值。 在数轴上表示 $3(x^2 - 5) > x - 1$ 的解集。 求整数 $p$ 的最大值，使得 $-2x^2 + x - p$ 对所有实数 $x$ 都有实根。

1. Form a quadratic equation by equating the line and the curve

From the line equation, $y = 8 - 2x$.
Substitute this into the curve equation:
$8 - 2x = 2x^2 - 6x + c$
$2x^2 - 4x + c - 8 = 0$

2. Apply the discriminant condition for a tangent

For the line to be a tangent to the curve, the quadratic equation must have equal roots, so the discriminant $b^2 - 4ac = 0$.
$(-4)^2 - 4(2)(c - 8) = 0$

3. Solve for $c$

$16 - 8(c - 8) = 0$
$16 - 8c + 64 = 0$
$80 - 8c = 0$
$8c = 80$
$c = 10$

1. Expand and rearrange the inequality into standard quadratic form

$3(x^2 - 5) > x - 1$
$3x^2 - 15 > x - 1$
$3x^2 - x - 14 > 0$

2. Factorise the quadratic expression to find critical values

$(3x - 7)(x + 2) > 0$
Critical values are $x = \frac{7}{3}$ and $x = -2$.

3. Determine the solution set and represent it on a number line

Since the inequality is $> 0$, the solution is $x < -2$ or $x > \frac{7}{3}$.

(Draw a number line with open circles at $-2$ and $\frac{7}{3}$, with an arrow pointing left from $-2$ and an arrow pointing right from $\frac{7}{3}$.)

1. Identify coefficients and write the discriminant condition for real roots

For $-2x^2 + x - p = 0$: $a = -2$, $b = 1$, $c = -p$.

For real roots, $b^2 - 4ac \geq 0$.

2. Substitute and simplify the discriminant inequality

$1^2 - 4(-2)(-p) \geq 0$
$1 - 8p \geq 0$
$p \leq \frac{1}{8}$

3. Select the greatest integer value of p

Since $p \leq \dfrac{1}{8} = 0.125$, the greatest integer value is
$p = 0.$