Using the Quadratic Formula for a question involving 2 variables

k isn't really a variable; it's a constant whose value is unknown.
A quadratic equation has one real root when the discriminant (value under the square root sign) is 0.
It has two real roots when the discriminant is positive.

The discriminant is
k² - 4 * 6 * 2
= k² - 48

For one real root:
k² - 48 = 0
k² = 48
k = sqrt(48) = sqrt(16) * sqrt(3) = 4 * sqrt(3)

For two real roots:
k² - 48 > 0
k² >=48
k > 4* sqrt(3) or k < -4 * sqrt(3)

First you set up the quadratic equation.

( -k ± √( k^2 - 4(6)(2) ) / 2(6)


If the determinant = 0, there is only 1 real solution. So you set the determinant equal to zero and solve for k:

k^2 - 4(6)(2) = 0
k^2 - 48 = 0
k^2 = 48
k =√(48)

If the determinant is > 0, there are two real solutions. So you set the determinant > than zero and solve for k:

k^2 - 48 > 0
k^2 > 48
k > √48

For a quadratic in the standard form ax^2+bx+c, it has one real root if b^2-4ac=0 and two real roots if b^c-4ac>0.

So just substitute in values

b^2-4ac=0
k^2-4*6*2=0
k^2-48=0
k^2=48
k=plus/minus sqrt(48)
k=plus/minus 4sqrt(3)

b^2-4ac>0
k^2-4*6*2>0
k^2-48>0
k^2>48
k>sqrt(48) and k<-sqrt(48)

6x^2 + kx + 2 = 0

x = [-k + or - square root(k^2 - 4*6*3)]/(2*6)

x will have only 1 root when square root (k^2 - 4*6*3) = 0
k^2 = 48
k = 4root (3)
--------------------------

x will have 2 real roots when (k^2 - 4*6*3) > 0
k > 4root(3)