Question

For what value(s) of b will cause 6x^2+bx-7=0 to have two real number solutions

Answer 1

B can be All real values.

Explanation:

The discriminant = b^2 - 4*6*-7 > 0

b^2 + 168 > 0

b^2 > -168

All values of b will be > 0.