Solve the wave equation, and the binomial function.

The general solution to this equation is given by d'Alembert's formula, which represents the superposition of two waves traveling in opposite directions.

where $f$ and $g$ are arbitrary twice-differentiable functions.
The general solution is $\boxed{y(x,t) = f(x - ct) + g(x + ct)}$.

Comparing this to $3x^2 + 5x - 12$:
For the $x^2$ term: $2a = 3 \implies a = \frac{3}{2}$
For the $x$ term: $2b = 5 \implies b = \frac{5}{2}$
For the constant term: $2a + 2c = -12$
Substitute $a = \frac{3}{2}$: $2\left(\frac{3}{2}\right) + 2c = -12 \implies 3 + 2c = -12 \implies 2c = -15 \implies c = -\frac{15}{2}$

Using the values $a = \frac{3}{2}$, $b = \frac{5}{2}$, and $c = -\frac{15}{2}$:

The function is $\boxed{F(x) = \frac{3}{2}x^2 + \frac{5}{2}x - \frac{15}{2}}$.

• $F(2) = \frac{3}{2}(2)^2 + \frac{5}{2}(2) - \frac{15}{2} = \frac{3}{2}(4) + 5 - \frac{15}{2} = 6 + 5 - \frac{15}{2} = 11 - \frac{15}{2} = \frac{22-15}{2} = \boxed{\frac{7}{2}}$
• $F(3) = \frac{3}{2}(3)^2 + \frac{5}{2}(3) - \frac{15}{2} = \frac{3}{2}(9) + \frac{15}{2} - \frac{15}{2} = \boxed{\frac{27}{2}}$
• $F(-2) = \frac{3}{2}(-2)^2 + \frac{5}{2}(-2) - \frac{15}{2} = \frac{3}{2}(4) - 5 - \frac{15}{2} = 6 - 5 - \frac{15}{2} = 1 - \frac{15}{2} = \frac{2-15}{2} = \boxed{-\frac{13}{2}}$
• $F(-3) = \frac{3}{2}(-3)^2 + \frac{5}{2}(-3) - \frac{15}{2} = \frac{3}{2}(9) - \frac{15}{2} - \frac{15}{2} = \frac{27}{2} - \frac{30}{2} = \boxed{-\frac{3}{2}}$

Since the coefficient of $x^2$ is $a = \frac{3}{2}$, which is positive ($a > 0$), the parabola opens upwards. Therefore, $F(x)$ has a minimum value.
The x-coordinate of the vertex (the minimum point) is given by $x = -\frac{b}{2a}$.

Now, substitute this x-value back into $F(x)$ to find the y-coordinate (the minimum value):

The function $F(x)$ has a minimum value. The coordinates of the minimum point are $\boxed{\left(-\frac{5}{6}, -\frac{205}{24}\right)}$.