CPM Homework Help : A2C Problem 9-110

You are given the equation $5x^{2} + bx + 20 = 0$ . For what values of $b$ does this equation have real solutions?

Hint:

Use the Quadratic Formula.

$x=\frac{-b\pm\sqrt{b^{2}-4(5)(20)}}{2(5)}$

More Help:

The equation will have real solutions as long as the value under the radical is greater than or equal to zero.