Find 2 missing numbers in an array of integers with two missing values

Question

How do you do this? The values are unsorted but are of [1..n] Example array [3,1,2,5,7,8]. Answer: 4, 6

I saw this solution in

Answer 1

#include 
#include 

/*
    the sum should be 1+...+n = n(n+1)/2
    the sum of squares should be 1^2+...+n^2 = n(n+1)(2n+1)/6.
*/

void find_missing_2_numbers(int *arr, int n);

int main()
{
    int arr[] = {3,7,1,6,8,5};

    find_missing_2_numbers(arr, 8);

    return 0;
}

void find_missing_2_numbers(int *arr, int n)
{

    int i, size, a, b, sum, sum_of_sqr, a_p_b, as_p_bs, a_i_b, a_m_b;
    size = n - 2;

    sum = 0;
    sum_of_sqr = 0;
    for (i = 0; i < size; i++)
    {
        sum += arr[i];
        sum_of_sqr += (arr[i] * arr[i]);
    }

    a_p_b = (n*(n+1))/2 - sum;
    as_p_bs = (n*(n+1)*(2 * n + 1))/6 - sum_of_sqr;
    a_i_b = ((a_p_b * a_p_b) - as_p_bs ) / 2;
    a_m_b = (int) sqrt((a_p_b * a_p_b) - 4 * a_i_b);
    a = (a_p_b + a_m_b) / 2;
    b = a_p_b - a;

    printf ("A: %d, B: %d\n", a, b);
}