help solve please?

Question

Solve 6x2 + 3x – 18 = 0 using the quadratic formula.

Answer 1

x= {-3 +/- (sq.rt) 3*3 - 4(6*-18)} /2*6
x= {-3 +/- (sq.rt) 9 - 432} /12
x= {-3 + 21} /12, {-3 - 21}/12
x= 1.5, -2

Answer 2

Here's the quadratic formula:
x = -B +/- √(B^2 - 4AC
Where A, B, C are the coefficients as shown:
Ax^2 + Bx + C = 0
In this case, A is 6, B is 3, and C is -18
Plugging in the formula, we get:
(-3 +/- √(3^2 - 4(6)(-18)))/2(6) =
(-3 +/- √(9 + 432))/12 =
(-3 +/- 21)/12 =
18/12, -24/12 =
3/2, -2

Answer 3

Im sure you are familiar with the quadratic formula so i will cut to the chase:

-3+/-sqrt of (9-4*6*-18) all that divided by 2*6 btw 2*6 means two times six (just to be on the safe side)

then -3+/- 21 all divided by 12

your two answers are 3/2 and -2

Answer 4

first off, to denote "squarred" the correct useage is (Shift + 6) to make a ^

6x^2 + 3x -18 = 0
these all have a three in common
3(2x^2 + 1x -6) =0
use the master product rule to find two digits that when mutliplied together are -12 and when added are +1
3(2x^2 + 4x -3x -6)
group the middle terms (reverse foil)
3[2x(x+2) -3(x+2)] when two of these match you are on the right track!
3[(2x-3)(x+2)]=0
so 2x-3 = 0 or x+2=0
2x-3=0
2x=3
x=3/2

OR

x+2=0
x= -2

solution set is {-2, 2/3}

Answer 5

x=2
don't know what a quadratic formula is - but 3x = 6