Need some help here.

Okay, math nerds.

I'm trying to solve this math problem. I'm trying to tutor someone, and this problem is really stumping me. Any help?

3x^2 – 2x + 7 = 0, then (x-1/3)^2 = ?

I need the answer in a fraction, and how you arrived at that answer.

First, use the quadratic formula to find the solutions, and then substitute the answers into (x-1/3)^2 to find the answer.

The quadratic gives me

[2 +- sqrt(-80)]/6]

which is imaginary. I'm starting to think maybe the question was misprinted.

No problems, if it is an imaginary root, then use this imaginary root.

Alright. Let's use [2 + sqrt(-80)]/6 first. That's

[2 + 4sqrt(-5)]/6

Or

2/6 + 4sqrt(-5)/6

Or

1/3 + 2sqrt(-5)/3

Plugged into the second equation, that's

[1/3 + 2sqrt(-5)/3 - 1/3]^2

Which is

[2sqrt(-5)/3]^2

Which is (4*-5)/9, or -20/9, which is one of the answers on the sheet. Holy crap, that took forever. Thanks.

>I'm trying to tutor someone

It's been a couple of years. I'm rusty. Right now we're covering adding and subtracting fractions. I'm getting familiar with the difficult stuff now, so I can be prepared when we get to it.

>I'm getting familiar with the difficult stuff now
lol wut

>>7
>difficult

Lol idiot. Can't even do any geodesics.

>>5

Here's a much easier solution. 

We want to know (x - 1/3)^2.
Let's call this y.  We want to find y.  Then:

y = (x - 1/3)^2

y = x^2 - 2/3 x + 1/9

3y = 3x^2 - 2x + 1/3

3y + 20/3 = 3x^2 - 2x + 7

But the right-hand side of this last equation was given to be zero.  So we have:

3y + 20/3 = 0

 y        = -20/9

Do I get the prize?