Here is the parabola and the tangent line we want:



We solve this system to find their common point:



by substitution:

4 - x² = mx + 5

-x² - mx - 1 = 0

 x² + mx + 1 = 0

This must have exactly ONE real solution, for if it has
TWO real solutions the line will cut the parabola in TWO
points, and not be tangent, like this:



Or if it has NO real solutions but only imaginary solutions
it will not touch the parabola at all, like this:


  
So to get it tangent to the parabola, we must be sure that

 x² + mx + 1 = 0

has only ONE solution for x.  That means its discriminant b²-4ac must
equal to 0, so

a = 1, b = m, c = 1

discriminant = b² - 4ac = m² - 4(1)(1) = m² - 4

This discriminant must = 0, so

m² - 4 = 0

(m - 2)(m + 2) = 0

m - 2 = 0;  m + 2 = 0
    m = 2;      m = -2

So there are two possible values for m, which means there are TWO lines
of the form y = mx + 5 which will be tangent to the parabola.  They are

y = 2x + 5 and y = -2x + 5



Answer: m = 2,  m = -2

Edwin