Question 255667
42. Find b^2 - 4ac and the number of real solutions to each equation.
If you have a quadrate: y = ax^2 + bx + c
The roots are x = [-b +- sqrt(b^2-4ac)]/(2a)
---
If b^2 - 4ac = 0 the two roots are both x = -b/2a
---
If b^2 - 4ac > 0 you will have two Real Number solutions
---
If b^2 -4ac < 0 you will get two Complex Number solutions
----------------------------------------------------------------- 
14. Solve each equation by using the quadratic formula. 
-8q^2+2q+1=0
x = [-2 +- sqrt(4 - 4*-8*1)]/(-16)
---
x = [-2 +- sqrt(36)]/(-16)
---
x = [-2 +- 6]/(-16)
---
x = [-8/-16] or x = [4/-16]
x = 1/2 or x = -1/4
=======================
Cheers,
Stan H.