Posted by **Lexie** on Thursday, April 22, 2010 at 10:25pm.

Solve the quadratic equation.

3x^2-15x+18=0

I know that the formula to quadratic equation is x=-b+and-(square root)b^2-4ac divided by 2a.

I know that a=3, b=-15 and c=18, then subsitute in and get two answers, my problem is that I keep getting an answer that does not work after I subsitute in the original equation. What do I do?

- Algebra -
**Reiny**, Thursday, April 22, 2010 at 10:52pm
did you get

x = (15 ± √(225 - 4(3)(18) )/6 ?

= (15 ± √9)/6

= (15 ±3)/6

= 3 or 2

we could have divided the original equation by 3 to get

x^2 - 5x + 6 = 0

now it factors

(x-3)(x-2) = 0

so x = 3 or x = 2

- Algebra -
**Lexie**, Thursday, April 22, 2010 at 11:01pm
I got 3 and -2. But to check it, wouldn't you substitue it the 3 and then the -2 in the original equation and it should be 0=0?

- Algebra -
**Reiny**, Friday, April 23, 2010 at 6:46am
and so it does

How did you get -2?

if you had a + and a - then you couldn't possible end up with +18 at the end.

if x= 3

3(9) - 45 + 18 = ??

if x=2

3(42) - 30 + 18 = ???

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