Solve the equation (11−3x)2 = 4(x+4)(x−1)

The given equation is $(11-3x)^2 = 4(x+4)(x-1)$.

Expand the left side $(11-3x)^2$ using the formula $(a-b)^2 = a^2 - 2ab + b^2$:

$$(11-3x)^2 = 11^2 - 2(11)(3x) + (3x)^2$$ $$= 121 - 66x + 9x^2$$

Expand the right side $4(x+4)(x-1)$. First, multiply the binomials $(x+4)(x-1)$:

$$(x+4)(x-1) = x(x-1) + 4(x-1)$$ $$= x^2 - x + 4x - 4$$ $$= x^2 + 3x - 4$$

Now, multiply by 4:

$$4(x^2 + 3x - 4) = 4x^2 + 12x - 16$$

Subtract $4x^2$, $12x$, and $-16$ from both sides to move all terms to one side:

$$9x^2 - 4x^2 - 66x - 12x + 121 + 16 = 0$$ $$5x^2 - 78x + 137 = 0$$

Here, $a=5$, $b=-78$, and $c=137$.

Calculate the discriminant $\Delta = b^2 - 4ac$:

$$\Delta = (-78)^2 - 4(5)(137)$$ $$\Delta = 6084 - 20(137)$$ $$\Delta = 6084 - 2740$$ $$\Delta = 3344$$

Now, substitute the values into the quadratic formula:

$$x = \frac{-(-78) \pm \sqrt{3344}}{2(5)}$$ $$x = \frac{78 \pm \sqrt{3344}}{10}$$

Factorize 3344 to find any perfect square factors:

$$3344 = 16 \times 209$$

So, $\sqrt{3344} = \sqrt{16 \times 209} = \sqrt{16} \times \sqrt{209} = 4\sqrt{209}$.

Substitute this back into the expression for $x$:

$$x = \frac{78 \pm 4\sqrt{209}}{10}$$

$$x = \frac{2(39 \pm 2\sqrt{209})}{10}$$ $$x = \frac{39 \pm 2\sqrt{209}}{5}$$

The solutions for $x$ are:

$$x_1 = \frac{39 + 2\sqrt{209}}{5}$$ $$x_2 = \frac{39 - 2\sqrt{209}}{5}$$

The solutions are \boxed{\text{x = \frac{39 \pm 2\sqrt{209}}{5}}}.