Question 1108399
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The solution by the other tutor is a perfectly good one, using a standard algebraic technique.  Here is another method that is often useful on problems like this.<br>
In any quadratic sequence, the row of "second differences" is constant.  Here is an example.<br>
For f(x) = 2x^2+3x-4, the first few terms of the sequence are
1, 10, 23, 40, 61
The "first differences" (differences between successive terms of the sequence) ar
9, 13, 17, 21
The "second differences" (differences between successive first differences) are
4, 4, 4<br>
This will always be the case for a quadratic sequence.<br>
So we can apply this knowledge to solving your problem.<br>
The terms of your sequence are
-3, -2, x, -6, y
The first differences are
1, x+2, -6-x, y+6
The second differences are
x+1, -2x-8, x+y+12<br>
Since we know the second differences are constant, we have
{{{x+1 = -2x-8}}}
{{{3x = -9}}}
{{{x = -3}}}<br>
Then
{{{-2x-8 = x+y+12}}}
{{{6-8 = -3+y+12}}}
{{{-2 = y+9}}}
{{{-11 = y}}}<br>
As found by the other tutor, the missing numbers are x = -3 and y = -11.