Solve the quadratic equations using the quadratic formula and completing the square. Simplify the surds.

$$
x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(2)}}{2(3)}
$$
$$
x = \frac{7 \pm \sqrt{49 - 24}}{6}
$$
$$
x = \frac{7 \pm \sqrt{25}}{6}
$$
$$
x = \frac{7 \pm 5}{6}
$$

$$
x_1 = \frac{7 + 5}{6} = \frac{12}{6} = 2
$$
$$
x_2 = \frac{7 - 5}{6} = \frac{2}{6} = \frac{1}{3}
$$
The solutions are $x=2$ and $x=\frac{1}{3}$.
$$
\boxed{x = 2, x = \frac{1}{3}}
$$

$$
x^2 + 6x = -5
$$

$$
x^2 + 6x + 9 = -5 + 9
$$

$$
(x+3)^2 = 4
$$

$$
x+3 = \pm\sqrt{4}
$$
$$
x+3 = \pm 2
$$

$$
x = -3 \pm 2
$$
$$
x_1 = -3 + 2 = -1
$$
$$
x_2 = -3 - 2 = -5
$$
The solutions are $x=-1$ and $x=-5$.
$$
\boxed{x = -1, x = -5}
$$

$$
\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}
$$
$$
\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}
$$

$$
5\sqrt{2} + 3(2\sqrt{2})
$$
$$
5\sqrt{2} + 6\sqrt{2}
$$

$$
11\sqrt{2}
$$
The simplified expression is $11\sqrt{2}$.
$$
\boxed{11\sqrt{2}}
$$

$$
2^{3 + (-5)}
$$
$$
2^{3 - 5}
$$
$$
2^{-2}
$$

$$
\frac{1}{2^2}
$$
$$
\frac{1}{4}
$$
The simplified expression is $\frac{1}{4}$.
$$
\boxed{\frac{1}{4}}
$$

$$
\log_{10}(100) = \log_{10}(10^2) = 2
$$
$$
\log_{10}(0.001) = \log_{10}(10^{-3}) = -3
$$

$$
2 + (-3) = -1
$$
The simplified expression is $-1$.
$$
\boxed{-1}
$$

$$
x = 2^5
$$

$$
x = 32
$$
The solution is $x=32$.
$$
\boxed{x = 32}
$$

$$
x = \arcsin\left(\frac{1}{2}\right) = 30^\circ
$$

$$
x = 180^\circ - 30^\circ = 150^\circ
$$
The solutions in the given range are $30^\circ$ and $150^\circ$.
$$
\boxed{x = 30^\circ, 150^\circ}
$$

$$
\sin x = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{h}
$$
$$
\cos x = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{a}{h}
$$

$$
\sin^2 x + \cos^2 x = \left(\frac{y}{h}\right)^2 + \left(\frac{a}{h}\right)^2
$$
$$
\sin^2 x + \cos^2 x = \frac{y^2}{h^2} + \frac{a^2}{h^2}
$$
$$
\sin^2 x + \cos^2 x = \frac{y^2 + a^2}{h^2}
$$

$$
\sin^2 x + \cos^2 x = \frac{h^2}{h^2}
$$
$$
\sin^2 x + \cos^2 x = 1
$$
The identity is proven.
$$
\boxed{\sin^2 x + \cos^2 x = 1 \text{ (Proven)}}
$$

$$
\frac{dy}{dx} = \frac{d}{dx}(4x^3) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(2)
$$
$$
\frac{dy}{dx} = 4(3)x^{3-1} - 3(2)x^{2-1} + 0
$$
$$
\frac{dy}{dx} = 12x^2 - 6x
$$
The derivative is $12x^2 - 6x$.
$$
\boxed{\frac{dy}{dx} = 12x^2 - 6x}
$$

$$
\frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(3x)
$$
$$
\frac{dy}{dx} = 2x^{2-1} + 3x^{1-1}
$$
$$
\frac{dy}{dx} = 2x + 3
$$

$$
\frac{dy}{dx}\Big|_{x=1} = 2(1) + 3
$$
$$
\frac{dy}{dx}\Big|_{x=1} = 2 + 3
$$
$$
\frac{dy}{dx}\Big|_{x=1} = 5
$$
The gradient at $x=1$ is 5.
$$
\boxed{\text{Gradient} = 5}
$$

$$
\int 3x^2\,dx + \int 2x\,dx
$$
$$
3\left(\frac{x^{2+1}}{2+1}\right) + 2\left(\frac{x^{1+1}}{1+1}\right) + C
$$
$$
3\left(\frac{x^3}{3}\right) + 2\left(\frac{x^2}{2}\right) + C
$$

$$
x^3 + x^2 + C
$$
The integral is $x^3 + x^2 + C$.
$$
\boxed{x^3 + x^2 + C}
$$

$$
\text{Area} = \int_0^5 x\,dx
$$

$$
\int x\,dx = \frac{x^{1+1}}{1+1} = \frac{x^2}{2}
$$

$$
\text{Area} = \left[\frac{x^2}{2}\right]_0^5
$$
$$
\text{Area} = \frac{5^2}{2} - \frac{0^2}{2}
$$
$$
\text{Area} = \frac{25}{2} - 0
$$
$$
\text{Area} = \frac{25}{2}
$$
The area under $y=x$ from 0 to 5 is $\frac{25}{2}$ square units.
$$
\boxed{\text{Area} = \frac{25}{2} \text{ square units}}
$$