Quadratic Equations

$$
3x^2 - x - 6x + 2 = 0
$$

$$
x(3x - 1) - 2(3x - 1) = 0
$$

$$
(3x - 1)(x - 2) = 0
$$

$$
3x - 1 = 0 \implies 3x = 1 \implies x = \frac{1}{3}
$$
$$
x - 2 = 0 \implies x = 2
$$
The solutions are $\boxed{x = \frac{1}{3}, x = 2}$.

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

Half of 6 is 3, and $3^2 = 9$.
$$
x^2 + 6x + 9 = -5 + 9
$$

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

$$
x + 3 = 2 \implies x = 2 - 3 \implies x = -1
$$
$$
x + 3 = -2 \implies x = -2 - 3 \implies x = -5
$$
The solutions are $\boxed{x = -1, x = -5}$.

$$
\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}
$$
$$
3\sqrt{8} = 3\sqrt{4 \times 2} = 3 \times \sqrt{4} \times \sqrt{2} = 3 \times 2 \times \sqrt{2} = 6\sqrt{2}
$$

$$
5\sqrt{2} + 6\sqrt{2} = (5 + 6)\sqrt{2} = 11\sqrt{2}
$$
The simplified expression is $\boxed{11\sqrt{2}}$.

$$
2^3 \times 2^{-5} = 2^{3 + (-5)}
$$

$$
2^{3 - 5} = 2^{-2}
$$

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

For $\log_{10}(100)$: We ask, "To what power must 10 be raised to get 100?"
$$
10^2 = 100 \implies \log_{10}(100) = 2
$$
For $\log_{10}(0.001)$: We ask, "To what power must 10 be raised to get 0.001?"
$$
0.001 = \frac{1}{1000} = \frac{1}{10^3} = 10^{-3}
$$
$$
\log_{10}(0.001) = -3
$$

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

$$
2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32
$$
The solution is $\boxed{x = 32}$.