Let's solve the given equation step by step:

\[
(2x-1)^2 - (x+)(x4) = 1 - 3(x-5)
\]

First, let's correct the apparent typo in the equation. It seems like \((x+)(x4)\) should be \((x+4)(x)\):

\[
(2x-1)^2 - (x+4)(x) = 1 - 3(x-5)
\]

Now, let's expand and simplify each part of the equation.

1. Expand \((2x-1)^2\):

\[
(2x-1)^2 = (2x-1)(2x-1) = 4x^2 - 4x + 1
\]

2. Expand \((x+4)(x)\):

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

3. Expand \(1 - 3(x-5)\):

\[
1 - 3(x-5) = 1 - 3x + 15 = 16 - 3x
\]

Now, substitute these expanded forms back into the original equation:

\[
4x^2 - 4x + 1 - (x^2 + 4x) = 16 - 3x
\]

Simplify the left side of the equation:

\[
4x^2 - 4x + 1 - x^2 - 4x = 16 - 3x
\]

Combine like terms:

\[
3x^2 - 8x + 1 = 16 - 3x
\]

Move all terms to one side of the equation to set it to zero:

\[
3x^2 - 8x + 1 - 16 + 3x = 0
\]

Combine like terms again:

\[
3x^2 - 5x - 15 = 0
\]

Now we have a quadratic equation:

\[
3x^2 - 5x - 15 = 0
\]

To solve this quadratic equation, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -5\), and \(c = -15\).

First, calculate the discriminant:

\[
b^2 - 4ac = (-5)^2 - 4(3)(-15) = 25 + 180 = 205
\]

Now, apply the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{205}}{2(3)} = \frac{5 \pm \sqrt{205}}{6}
\]

So, the solutions to the equation are:

\[
x = \frac{5 + \sqrt{205}}{6} \quad \text{and} \quad x = \frac{5 - \sqrt{205}}{6}
\]