We are given the function $f(x) = x^2 - 3x + 4$. We want to find the values of $x$ such that $f(x) = f(2x+1)$.

1. Problem Description

We are given the function $f(x) = x^2 - 3x + 4$. We want to find the values of $x$ such that $f(x) = f(2x+1)$.

2. Solution Steps

We are given $f(x) = x^2 - 3x + 4$.

We need to find the values of $x$ that satisfy the equation $f(x) = f(2x+1)$.

Substituting the expression for $f(x)$ into the equation, we have:

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

Expanding the terms, we get:

$x^2 - 3x + 4 = (4x^2 + 4x + 1) - (6x + 3) + 4$

$x^2 - 3x + 4 = 4x^2 + 4x + 1 - 6x - 3 + 4$

$x^2 - 3x + 4 = 4x^2 - 2x + 2$

Now, we rearrange the equation to set it equal to zero:

$0 = 4x^2 - x^2 - 2x + 3x + 2 - 4$

$0 = 3x^2 + x - 2$

We can factor this quadratic equation:

$3x^2 + x - 2 = 0$

$3x^2 + 3x - 2x - 2 = 0$

$3x(x+1) - 2(x+1) = 0$

$(3x - 2)(x + 1) = 0$

Therefore, either $3x - 2 = 0$ or $x + 1 = 0$.

If $3x - 2 = 0$, then $3x = 2$, so $x = \frac{2}{3}$.

If $x + 1 = 0$, then $x = -1$.

Thus, the values of $x$ that satisfy the equation are $x = \frac{2}{3}$ and $x = -1$.

3. Final Answer

The values of $x$ that satisfy the equation $f(x) = f(2x+1)$ are $x = \frac{2}{3}$ and $x = -1$.