Let's solve each equation step by step.

1. Solve \[x^3 - 16x = 0\]

Factor the equation as follows:

\[
x(x^2 - 16) = 0
\]

This gives us two factors:

\[
x = 0 \quad \text{or} \quad x^2 - 16 = 0
\]

Solve \(x^2 - 16 = 0\):

\[
x^2 = 16 \implies x = \pm 4
\]

Therefore, the solutions are \(x = 0\), \(x = 4\), and \(x = -4\).

2. Solve \[x^2 + 3x - 10 = 0\]

Use the quadratic formula: \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\), where \(a=1\), \(b=3\), and \(c=-10\).

\[
x = \frac{{-3 \pm \sqrt{{3^2 - 4 \times 1 \times (-10)}}}}{2 \times 1} = \frac{{-3 \pm \sqrt}}{2}
\]

\[
x = \frac{{-3 \pm \sqrt{49}}}{2} = \frac{{-3 \pm 7}}{2}
\]

\[
x = 2 \quad \text{or} \quad x = -5
\]

3. Solve \[3x^2 + 5x + 2 = 0\]

Apply the quadratic formula:

\[
x = \frac{{-5 \pm \sqrt{{5^2 - 4 \times 3 \times 2}}}}{2 \times 3}
\]

\[
x = \frac{{-5 \pm \sqrt}}{6} = \frac{{-5 \pm 1}}{6}
\]

\[
x = -\frac{4}{6} = -\frac{2}{3} \quad \text{or} \quad x = -1
\]

4. Solve \((x-1)^2 + x(4-x) = 11\)

Expand and simplify:

\[
(x-1)^2 = x^2 - 2x + 1
\]

\[
x(4-x) = 4x - x^2
\]

Combine both and set equal to 11:

\[
x^2 - 2x + 1 + 4x - x^2 = 11
\]

\[
2x + 1 = 11 \implies 2x = 10 \implies x = 5
\]

5. Solve \((x-5)^2 - x(x+2) = 5\)

Expand and simplify:

\[
(x-5)^2 = x^2 - 10x + 25
\]

\[
x(x+2) = x^2 + 2x
\]

Set the equations together:

\[
x^2 - 10x + 25 - x^2 - 2x = 5
\]

\[
-12x + 25 = 5 \implies -12x = -20 \implies x = \frac{5}{3}
\]

6. Solve \(x(x+4) - x^2 - 6x = 10\)

Expand and simplify:

\[
x^2 + 4x - x^2 - 6x = 10
\]

\[
-2x = 10 \implies x = -5
\]

7. Solve \(6x^2 - (2x-3)(3x+2) = 1\)

Expand the term:

\[
(2x-3)(3x+2) = 6x^2 + 4x - 9x - 6 = 6x^2 - 5x - 6
\]

Substitute back:

\[
6x^2 - (6x^2 - 5x - 6) = 1
\]

\[
5x + 6 = 1 \implies 5x = -5 \implies x = -1
\]

8. Solve \((x+2)(x-5) + x(1-x) = 2\)

Expand and simplify:

\[
(x+2)(x-5) = x^2 - 5x + 2x - 10 = x^2 - 3x - 10
\]

\[
x(1-x) = x - x^2
\]

Substitute into the equation:

\[
x^2 - 3x - 10 + x - x^2 = 2
\]

\[
-2x - 10 = 2 \implies -2x = 12 \implies x = -6
\]

9. Solve \((x-1)(2-x) + (x-3)^2 = 4-2x\)

Expand and simplify:

\[
(x-1)(2-x) = 2x - x^2 - 2 + x = -x^2 + 3x - 2
\]

\[
(x-3)^2 = x^2 - 6x + 9
\]

Substitute into the equation:

\[
-x^2 + 3x - 2 + x^2 - 6x + 9 = 4 - 2x
\]

\[
-3x + 7 = 4 - 2x \implies -x = -3 \implies x = 3
\]