Quadratic dilema

[adicus22]

I am completely stumped and don't know how to solve this for x even though I've reduced it down. Please help it is my last problem. If need be take it all the way back to square one and show me step by step where I went wrong if I have done so. Thank you very much, I have really appreciated the help I have been receiving from this forum. Keep it up.

Solve 6x2 + 3x – 18 = 0 using the quadratic formula.
Read the information in the assignment list to learn more about how to type math symbols, such as the square root.
Answer:
Show your work here: Below is my attempt at answering this.

x= (-3±?(3^2-4(6)(18)))/(2(6))

(-3±?(9 -432))/12

(-3±?(-423))/12

(-3±20.566)/12

 

[mmm4444bot]

… Read the information in the assignment list to learn more about how to type math symbols, such as the square root …

Why did you tell us this?

Hmmm. Are you using a calculator? If so, then you can use it to check your candidates for x, too!

Store a candidate value to x, and then enter the quadratic polynomial. If that value stored in x is a solution, then the calculator will display zero, right?

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x= (-3±?(3^2-4(6)(18)))/(2(6))

This is all good, except for one error.

C is not 18.

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±?(-423)

±20.566

We cannot take the square root of a negative value in the Real Number System.

Correct your value for c above; then you'll have some positive value inside the radical sign, instead.

 

[Denis]

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

Relax, buddy! Start by simplifying, dividing by 3:
2x^2 + x - 6 = 0

So you now have a=2, b=1 and c=-6 : give it another shot :idea:

 

[adicus22]

Thank you both very much for your time.