Please?
Find the constant $k$ such that the quadratic $2x^2 + 3x + k$ has a double root.
I'll mark the first actually helpful answer the brainliest. Just type the answer, I don't need an explanation...

  k = 9/8

Step-by-step explanation:

For a root with multiplicity 2, the discriminant must be zero. For quadratic ax^2 +bx +c, the discriminant is ...

  d = b^2 -4ac

For your quadratic, a=2, b=3, c=k, so you want ...

  3^2 -4(2)(k) = 0

  9 -8k = 0

  k = 9/8