Math help, Pre-Cal....?

Question

They told me something about having to use (b^2-4ac) to answer the problem but i have no clue on what to do, if you happen to know how to solve this problem please do so, with steps on how to do it please...

Find the discriminant, how many solutions does it have, real or imaginary. Then solve the quadratic equation:
4x^2-2x+9=0

Answer 1

The solution to a quadratic equation ax^2 + bx + c =0 is:
x = (- b +/- sqrt(b^2 - 4ac))/2a

Where the discriminant is the actually the terms inside the square root (b^2-4ac).

Given the quadratic equation: 4x^2-2x+9=0
a = 4
b = -2
c = 9

b^2 - 4ac = (-2)^2 - 4(4)(9) = 4 - 144 = -140 (discriminant)

The solutions are:

x =( -(-2) +/- sqrt(-140))/(2)(4) = (2 +/- sqrt(-140))/8
x = (2 +/- sqrt((4)(-35)) )/8

Answers (2 roots)
x = (2 + 2i sqrt(35))/8
x = (2 - 2i sqrt(35))/8

Answer 2

the discriminant deal is Algebra I stuff. Have you forgotten so soon?

Basically, if the discriminant is negative you will have complex roots
If it is zero, you have two rational roots numerically equal but with opposite signs
If it is a perfect square, you will have rational roots.
If it is not a perfect square, you will irrational roots.
If you don't know how to solve this in pre-calc, better bust open one of your old text books and find out.