Wondering if I answered this Question correctly

For what values of k will the equation 4x^2+kx+25=0 have real solutions.
I used the formula x= -b/2a and got k= 20 or k= -20
Thanks

To have real solutions, the discriminant must be non-negative:

b^2 - 4ac ≥ 0
k^2 - 4*4*25 ≥ 0
k^2 - 20^2 ≥ 0
(k - 20)(k + 20) ≥ 0

There are 3 intervals to consider: (-∞ , -20], [-20, 20], [20, ∞). Picking a point in [-20, 20], say 0 for easy calculation, we find this yields a negative discriminant. Thus, the answer is (-∞, -20], [20, ∞).

You're close, but you're not seeing the entire picture.

Using the Quadratic Formula:

[ -b +- sqrt( b^2 - 4ac ) ] / 2a
[ -k +- sqrt( k^2 - 4(4)(25) ) ] / 2(4)
[ -k +- sqrt( k^2 - 4(4)(25) ) ] / 8
[ -k +- sqrt( k^2 - 400 ) ] / 8

So in order to have real roots, the "k^2 - 400" part cannot be negative. That means:

k^2 - 400 >= 0 <----- Note >=, not just = like you did.
k^2 >= 400

So k >= 20, or k <= -20... or just |k| >= 20.