If the quadratic equation x^{2} + px + q = 0 has two solutions x_{1} and x_{2}, then x_{1} + x_{2} = - p and x_{1} x_{2} = q.
Note: p is the coefficient of the x term of the quadratic equation and q is the constant term of the quadratic equation.
For our case, x_{1} = 2 and x_{2} = -1/2. Then x_{1} + x_{2} = 2 + (-1/2) = 4/2 - 1/2 = 3/2. x_{1} x_{2} = 2 × (-1/2) = -1. Then the quadratic function is:
x^{2} - (x_{1} + x_{2})x + x_{1} x_{2} = 0. Substitute values into this quadratic equation, we get
x^{2} - (3/2)x + (-1) = 0. Multiply 2 to each term of the equation, we get 2x^{2} - 3x - 2 = 0. This is the quadratic equation that the question asked.

To check the result, we need to solve the quadratic equation to see whether the solution of 2x^{2} - 3x - 2 = 0 satisfy the given values. For the quadratic equation
2x^{2} - 3x - 2 = 0, a = 2, b = -3 and c = -2. Then b^{2} - 4ac = (-3)^{2} - 4(2)((-2) = 9 + 16 = 25 = 5^{2}. Then
x_{1} = [-b + square root of (b^{2} - 4ac)]/2a = [3 + 5]/[2 × 2] = 8/4 = 2. x_{2} = [-b - square root of (b^{2} - 4ac)]/2a = [3 - 5]/[2 × 2] = -2/4 = -1/2.
So its two solution are 2 and -1/2, which satisfy the given condition. So the quadratic equation is the solution.