Using the Quadratic Formula for a question involving 2 variables

This is the question exactly:

For what Values of k will the equation 6x^2 +kx +2 =0 have

a) One Real Root b) Two Real Roots

How would I do this?

Discriminant = b² - 4ac
(It should look familiar because it comes from the quadratic formula.)

The discriminant tells you the nature of your roots. If...
b² - 4ac > 0, there are two real roots
b² - 4ac = 0, there is one real root
b² - 4ac < 0, there are two complex roots.

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In the equation, 6x² + kx + 2 = 0,
a = 6; b = k; c = 2

Part A.
b² - 4ac = 0
k² - 4(6)(2) = 0
k² - 48 = 0
k² = 48
k = √48 or 4√3

Part B.
b² - 4ac > 0
Same thing. Solve for b.
k > 4√3

There will be one real root when the discriminant, the part of the quadratic formula that equals
sqrt(b^2 - 4ac) is 0, and two real roots when the discriminant is positive. You already know a = 6 and
c = 2, so put a = 6, b = k, and c = 2 into the discriminant.
For one real solution, use an equation, equate the discriminant with 0.
sqrt(k^2 - 4(6)(2)) = 0
Square both sides
k^2 - 4(6)(2) = 0
k^2 - 48 = 0
Factor
(k + 4sqrt(3))(k - 4sqrt(3)) = 0
Solve
One real root when k = 4sqrt(3) and when k = -4sqrt(3)

For two real solutions use an inequality since the discriminant has to be greater than 0
k^2 - 48 > 0
Since you can't solve for iequalities by factoring, you have to do this:
k^2 > 48
k < - 4sqrt(3) and k > 4sqrt(3) will give two real solutions.

Use the discriminant, b^2 - 4ac to determine what happens.
For this equation, its discriminant is k^2 - 48.
For a quadratic equation to have 1 real root, the discriminant will be equal to 0. If it has 2 real roots then the discriminant is more than 0.

a) k^2 - 48 = 0
k^2 = 48
k = sqrt(48), -sqrt(48) for the equation to have 1 real root.