Another Algebra II Question

I'm so stuck, and i have no idea where to start.. .

The question is : Write a quadratic equation with roots +/- The square root of 7, over 3.
Write the equation in the form ax^2 + bx + c = 0 , where a, b, and c, are integers.

this is simple when you think about it (word problems scare even the best) what it wants is and equation with -√7/3and +√7/3. to get your answer you must work backwords.
to begin put it in the equation x=-√7/3 and x=+√7/3
multiply both sides by 3 giving you 3x=+√7 and 3x=-√7
next move your √7 over giving 3x-√7 using your positive and 3x+√7 for your negative
then multply them together (3x-√7)(3x+√7) giving 9x^2 -7 as an answer ( you have no middle because both sets end in the same number w/ opposite signs.

so your answer is 9x^2-7

Think about where roots come from

x - 3 = 0.........note we use opposite operations
x= 3

2x + 5= 0......still using opposite operations
2x = -5
x = -5/2

(x + 5)^2 = 6..............take square root
x + 5 = ±√6
x = -5 ± √6

9x^2 - 7 = 0
9x^2 = 7
x^2 = 7/9....take square root
x = ±√7/3


so our polynomial is 9x^2 + 0x - 7 = 0
which becomes
9x^2 - 7 = 0