Question 79119
how do i decide all values of b in the following equations that will give one or more real number solutions? If an equation has solution how do i write a rule for judging by looking at it in standard form?

a. 3x^2+bx-3=0
b. 5x^2+bx+1=0
c. -3x^2+bx-3=0

----------
Rule:
b^2-4ac must be >=0
---------------
a. b^2-4*3*-3>=0
b^2>=-36
b can be any Real Number Value
.------------------------------------
b. b^2-4*5*3>=0
b^2>=60
b>=2sqrt15 or b<=-2sqrt15
------------------
c. b^2-4*-3*-3>=0
b^2>=36
b>=6 or b<=-6

=============
Cheers,
Stan H.