To find the zeros of the quadratic polynomial p(x) = 2x² + 7x – 4, we solve the equation 2x² + 7x – 4 = 0 using the quadratic formula: x = [– b ± √(b² – 4ac)] / (2a), where a = 2, b = 7, and c = – 4.

First, calculate the discriminant: D = b² – 4ac = 7² – 4(2)(– 4) = 49 + 32 = 81.

Now, solve for x: x = [– 7 ± √81] / (2*2) = [– 7 ± 9] / 4.

This gives two solutions: x = (– 7 + 9)/4 = 2/4 = 1/2 and x = (– 7 – 9)/4 = – 16/4 = – 4.

Therefore, the zeros are – 4 and 1/2.

Final Answer: $-4,\frac{1}{2}$