Given f(x) = x² – 3x + 4.

We need to find x satisfying f(x) = f(2x + 1).

We have f(2x + 1) = (2x + 1)² – 3(2x + 1) + 4

⇒ f(2x + 1) = (2x)² + 2(2x)(1) + 1² – 6x – 3 + 4

⇒ f(2x + 1) = 4x² + 4x + 1 – 6x + 1

∴ f(2x + 1) = 4x² – 2x + 2

Now, f(x) = f(2x + 1)

⇒ x² – 3x + 4 = 4x² – 2x + 2

⇒ 3x² + x – 2 = 0

⇒ 3x² + 3x – 2x – 2 = 0

⇒ 3x(x + 1) – 2(x + 1) = 0

⇒ (x + 1)(3x – 2) = 0

⇒ x + 1 = 0 or 3x – 2 = 0

⇒ x = –1 or 3x = 2

∴ x = –1 or

Thus, the required values of x are –1 and.