That is close

Let $\large y = \dfrac{2x^2 - 5x +3}{ 4x -m}$ $2x^2 - (4y +5)x + 3 + my = 0$ As x is real , $discriminant \geq 0$ $(4y + 5)^2 - 8(3+my) \geq 0$ $16y^2 + (40-8m)y +1 \geq 0$ A quadratic in y is non-negative for all values of y if coefficient of $y^2$ is positive and $discriminat \leq 0$ $(40-8m)^2 - 4*16*1 \leq 0$ $(5-m)^2 - 1 \leq 0$ $(m-6)(m-4) \leq 0$ $m \epsilon [4 , 6]$ Consider m = 4 $y = \dfrac{(2x-3)(x-1)}{4(x-1)}$ $y = \dfrac{2x-3}{4} , x \neq 1$ Since $x \neq 1 , y \neq \dfrac{-1}{4}$ Hence y cannot take all real values for $m = 4 .$ So, $m = 4$ is not acceptable . Similarly $m = 6$ is to be rejected . So for the given expression to take all real values , $m \in (4, 6 )$ . There is only 1 integer in this range.