1) To solve the equation 2x(x-5)-x(3+2x)=6, we can start by simplifying the equation:

2x(x-5) - x(3+2x) = 6
2x^2 - 10x - 3x - 2x^2 = 6
-13x = 6

Now, divide both sides of the equation by -13:

x = 6/-13
x = -6/13

Therefore, the solution to the equation is x = -6/13.

2) To solve the equation x^2-6x+9=0, we can use the quadratic formula:

x = (-b ± √(b^2-4ac)) / (2a)

In this case, a = 1, b = -6, and c = 9. Substituting these values into the quadratic formula:

x = (-(-6) ± √((-6)^2-4(1)(9))) / (2(1))
x = (6 ± √(36-36)) / 2
x = (6 ± √0) / 2
x = (6 ± 0) / 2
x = 6 / 2
x = 3

Therefore, the solution to the equation is x = 3.

3) To solve the equation x^2-x-6=0, we can factorize it:

(x-3)(x+2) = 0

Setting each factor equal to zero:

x-3 = 0 or x+2 = 0

Solving for x in each equation:

x = 3 or x = -2

Therefore, the solutions to the equation are x = 3 and x = -2.

4) To solve the equation x^2-15x-2=0, we can again use the quadratic formula:

x = (-b ± √(b^2-4ac)) / (2a)

In this case, a = 1, b = -15, and c = -2. Substituting these values into the quadratic formula:

x = (-(-15) ± √((-15)^2-4(1)(-2))) / (2(1))
x = (15 ± √(225+8)) / 2
x = (15 ± √233) / 2

Therefore, the solutions to the equation are x = (15 + √233) / 2 and x = (15 - √233) / 2.