I will assume you used your formula correctly to get the equations. I will simply show you how to use Gauss-Jordan Elimination to find a0, a1 and a2. But before I start, there are a few terms I need to define:
R1 = row 1, R2 = row 2, R3 = row 3
C1 = column1, C2 = column 2, C3 = column 3
R1 + R2 => R2 [add R1 to R2 to get a new R2]
R1 / -2 => R1 [multiply R1 by (-2) to get a new R1]
R2 x 3 + R1 => R1 [multiply R2 by 3, then add R1 get a new R1]
This will make more sense when I actually do the steps. Now let's begin..
First, put the equations in augmented matrix form (3 rows, 4 columns):
1 2 4 5
1 3 9 2
1 4 16 5
That's the easy part, unless you have a graphing calculator that has the RREF function (Reduced Row Echelon Form) under the MATRIX category. Assuming you don't...
1. Use row ops to get the first element (R1C1) to be = 1. Lucky for us it already is.
2. Use row ops to make all of the numbers underneath the 1 in R1 = 0:
1 2 4 5
1 3 9 2 (R1 - R2 => R2)
1 4 16 5 (R1 - R3 => R3)
1 2 4 5
0 -1 -5 3
0 -2 -12 0
3. Make the R2C2 element = 1, then make all other elements in C2 =0:
1 2 4 5
0 -1 -5 3 (R2 x -1 => R2)
0 -2 -12 0
1 2 4 5 (R2 x -2 + R1 => R1)
0 1 5 -3
0 -2 -12 0 (R2 x 2 + R3 => R3)
1 0 -6 11
0 1 5 -3
0 0 -2 -6
4. You're almost done. Make the R3C3 element = 1, and the rest of the elements in C2 = 0:
1 0 -6 11
0 1 5 -3
0 0 -2 -6 (R3 / -2 => R3)
1 0 -6 11 (R3 x 6 + R1 => R1)
0 1 5 -3 (R3 x -5 + R1 => R1)
0 0 1 3
1 0 0 29
0 1 0 -18
0 0 1 3
This yields an answer of:
a0 = 29
a1 = -18
a2 = 3
Check these numbers in all three of your equations to verify there weren't any oops-es along the way (there aren't, I already checked using RREF on my graphing calculator).
Now that you have your a's, you are on your own!