1) 5x+(2x+1)(x-3)=0
5x + 2x^2 - 6x + x - 3 = 0
Rearranging:
2x^2 + x - 3 = 0
The equation is now a quadratic equation. Solving for x:

2x^2 + x - 3 = 0
Using the quadratic formula:
x = (-1 ± √(1 + 24)) / 4
x = (-1 ± √25) / 4
x = (-1 ± 5) / 4
x = (-1 + 5) / 4 or x = (-1 - 5) / 4
x = 4 / 4 or x = -6 / 4
x = 1 or x = -1.5

Therefore, the solutions to the equation are x = 1 and x = -1.5

2) x^2-5=(x-5)(2x-1)
Expanding the right side:
x^2 - 5 = 2x^2 - x - 10x + 5
Rearranging:
x^2 - 5 = 2x^2 - 11x
Rearranging again:
2x^2 - 11x - x^2 + 5 = 0
x^2 - 11x + 5 = 0
The equation is now a quadratic equation. Solving for x:

Using the quadratic formula:
x = (11 ± √(121 - 415)) / 2
x = (11 ± √(121 - 20)) / 2
x = (11 ± √101) / 2
x = (11 ± 10.05) / 2
x = (21.05) / 2 or x = (0.95) / 2
x = 10.525 or x = 0.475

Therefore, the solutions to the equation are x ≈ 10.525 and x ≈ 0.475.