Given:
The equation is y = 2x^2 - 3x + 4.

To find the value of x when y is minimum, we need to find the vertex of the parabola represented by the equation. The vertex form of a parabola is given by y = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex.

Finding the vertex:
Comparing the given equation y = 2x^2 - 3x + 4 with the vertex form y = a(x - h)^2 + k, we can find the coordinates of the vertex.

From the given equation, we can see that a = 2, h = -(-3)/2a = 3/4a = 3/8, and k = 4.

Therefore, the coordinates of the vertex are (3/8, 4).

Determining the minimum value:
Since the coefficient of x^2 is positive (2 > 0), the parabola opens upwards, and the vertex represents the minimum point of the parabola.

So, when x = 3/8, y is minimum.

Calculating the value of x:
Since we need to find the value of x when y is minimum, we substitute x = 3/8 into the equation y = 2x^2 - 3x + 4.

y = 2(3/8)^2 - 3(3/8) + 4
= 2(9/64) - 9/8 + 4
= 18/64 - 9/8 + 4
= 9/32 - 9/8 + 4
= 9/32 - 36/32 + 128/32
= (9 - 36 + 128)/32
= 101/32

Therefore, when y is minimum, x = 3/8 and y = 101/32.

Conclusion:
The value of x when y is minimum in the equation y = 2x^2 - 3x + 4 is x = 3/8.