Find the value of k if the equation 3x^2+(k+2)x+3=0 has a double real root

Question

I know that double real root means -b-4ac>0 but I don't know where to go from there

Answer 1

3x² + (k+2)x + 3 = 0

 

Double real root when B² - 4AC = 0

 

So, (k+2)² - 4(3)(3) = 0

 

      k² + 4k -32 = 0

 

      (k + 8)(k - 4) = 0     k = -8 or 4