SOLUTION: I have the whole quadratic equation, just want to make sure that I am plugging in the numbers correctly. if I have {{{3x^2-6x+2=0}}} then, in my quadratic eq a=3,b=6, and c=2, is t

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: I have the whole quadratic equation, just want to make sure that I am plugging in the numbers correctly. if I have {{{3x^2-6x+2=0}}} then, in my quadratic eq a=3,b=6, and c=2, is t      Log On


   

Question 54275This question is from textbook Algebra for College Students
: I have the whole quadratic equation, just want to make sure that I am plugging in the numbers correctly. if I have 3x%5E2-6x%2B2=0 then, in my quadratic eq a=3,b=6, and c=2, is that correct? From there I should just follow through with the eq ? I am getting -2 sqrt 3 and it just doesn't look right. This question is from textbook Algebra for College Students

Found 2 solutions by Earlsdon, funmath:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, you don't have it quite right!
If your equation is: 3x%5E2-6x%2B2+=+0, then a = 3, b = -6, and c = 2
Now if you apply the quadratic formula x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a:
x+=+%28-%28-6%29%2B-sqrt%28%28-6%29%5E2-4%283%29%282%29%29%29%2F2%283%29
x+=+%286%2B-sqrt%2836-24%29%29%2F6
x+=+%286%2B-sqrt%2812%29%29%2F6%29 Simplifying this, you get:
x+=+%286%2F6%29%2B2sqrt%283%29%2F6 and x+=+%286%2F6%29-2sqrt%283%29%2F6
x+=+1%2Bsqrt%283%29%2F3 and x+=+1-sqrt%283%29%2F3

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
Standard Form of a quadratic equation is highlight%28ax%5E2%2Bbx%2Bc=0%29
3x%5E2-6x%2B2=0
a=3, b=-6, c=2 (You had left the - off of the b.)
The Quadratic Formula is:
highlight%28x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29%29
Plug your numbers into the formula:
x=%28-%28-6%29%2B-sqrt%28%28-6%29%5E2-4%283%29%282%29%29%29%2F%282%283%29%29
x=%286%2B-sqrt%2836-24%29%29%2F6
x=%286%2B-sqrt%2812%29%29%2F6
x=%286%2B-sqrt%284%29%2Asqrt%283%29%29%2F6We factor 12 into 4*3 because 4 is a perfect square. The square root of 4 is 2, but we can't take the square root of three.
x=%286%2B-2%2Asqrt%283%29%29%2F6 I can't fix this glitch...+/-is supposed to be in front of the 2.
x=2%283%2B-sqrt%283%29%29%2F%282%2A3%29The 2'2 cancel because 2/2=1.
x=cross%282%29%283%2B-sqrt%283%29%29%2F%28cross%282%29%2A3%29
highlight%28x=%283%2B-sqrt%283%29%29%2F3%29
Happy Calculating!!!