Can You Help Me Solve 6x2+3x-18=0 Using The Quadratic Formula Can You Work It Out With Me?

First of all, do the multiplying  and/or dividing first. Then the addition and subtraction. You need to figure out what X is.

I would factor 3 out of all of the coefficients first, just to make the arithmetic a little bit easier.
  6x^2 + 3x - 18 = 0
  2x^2 + x - 6 = 0
The quadratic equation says the solution to ax^2 + bx + c = 0 is
  x = (-b ±√(b^2 - 4ac))/(2a)
Your (reduced) equation has a=2, b=1, c=-6. Substituting these into the formula gives
  x = (-1 ±√(1^2 - 4(2)(-6)))/(2(2))
  = (-1 ±√(1+48))/4
  = (-1 ±√49)/4
  = (-1±7)/4
  = {-8, +6}/4    (-1-7 = -8, -1+7 = +6, so there are two quantities represented here)
  x = {-2, +3/2}    (either of these values is a solution for x)

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You can also solve this by factoring
  2x^2 + x - 6 = 0
  2x^2 + 4x - 3x - 6 = 0
  2x(x + 2) - 3(x + 2) = 0
  (2x-3)(x+2) = 0
  x = 3/2 or x = -2