$$\sqrt{x^2 - 4x + 13} + 1 = 2x$$$$\implies \sqrt{x^2 - 4x + 13} = 2x-1$$$$\implies x^2 - 4x + 13 = 4x^2-4x +1$$$$x^2=4, x_1=-2, x_2=2$$Plug into the original equation, we know that $x_1=-2$ is extranous solution. The solution of the original equation is $\boxed {x=2}$.

x^2 - 4x +13 = (2x - 1)^2 x^2 - 4x +13 = 4x^2 - 4x + 1 3x^2 = 12 x^2 = 4 => x = 2 ∵ 2x - 1 ≥ 0 x ≥ 0.5

We have the equation $\sqrt{x^2-4x+13}+1 = 2x$. Subtracting one to both sides, we get $\sqrt{x^2-4x+13} = 2x-1$. Squaring both sides to remove the square root, we reach $x^2-4x+13 = 4x^2-4x+1$. We add $4x$ to both sides to eliminate that from both sides. We subtract $x^2$ and we get $13 = 3x^2+1$. Subtracting 1 from both sides, we get $12 = 3x^2$. Dividing by 3 on both sides, we get $4 = x^2$, and taking the square root, we get $x = \pm 2$. However, when we substitute $-2$ to $x$, We have $x^2-4x+13$ to evaluate to $25$, and the square root is positive 5, so the left side is 6. However, the right side evaluates to $-4$, so $-2$ is extraneous. Therefore, the only solution to the equation is $\boxed{x = 2}$

$\sqrt{x^2 - 4x + 13} + 1 = 2x \Rightarrow (\sqrt{x^2 - 4x + 13})^2 = (2x - 1)^2 \Rightarrow x^2 - 4x + 13 = 4x^2 - 4x + 1 \Rightarrow 3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2,$ $\text{but } x = -2 \text{ is extraneous so our solution is only } x = \boxed{2}.$