Let . For what value of is there exactly one real value of such that ?

A quadratic formula, when graphed, becomes a parabola. It will have two values from every value of x except for the vertex.

So, the problem becomes:  find the vertex.

This can be done by graphing, by completing the square, or by using this formula:  x  =  -b / (2a).

Because you have the function in the form:  f(x)  =  ax² + bx + c   --->  a = 1, b = 4, c = -31

 --->    x  =  -b / (2a)   --->   x  =  -4 / (2·1)  =  -2

The x-value of the vertex occurs when  x = -2, now:   f(-2)  =  (-2)² + 4(-2) - 31  =  -35 

This value, -35, is the answer that you are looking for.