Question: find 5 consecutive integers such that 3 times the product of the 3rd and 4th, minus the the product of the 2nd and 4 less than the first, plus 3 times the 5th is 232. Variables are x, x +1, x + 2, x + 3, x + 4 ?
Let us rewrite your problem statement.
find 5 consecutive integers such that
3 times the product of the 3rd and 4th,
minus the product of the 2nd and 4 less than the first,
plus 3 times the 5th
is 232.
Variables are x, x +1, x + 2, x + 3, x + 4, or v1, v2, v3, v4, v5
Turning those statements above into mathematical expressions,
3 times the product of the 3rd and 4th, becomes 3*v3*v4
minus the product of the 2nd and (4 less than the first), becomes - v2*(v1-4)
plus 3 times the 5th becomes + 3*v5
is 232. becomes = 232
As a single expression, this is
3*v3*v4 - v2*(v1 - 4) + 3*v5 = 232
3(x+2)(x+3) - (x+1)(x - 4) + 3(x+4) = 232
3(x^2 + 5x + 6) - (x^2 - 3x - 4) + (3x + 12) = 232
2x^2 + 21x + 34 = 232
2x^2 + 21x - 198 = 0
(2x + 33)(x - 6) = 0
x = -33/2, x = 6
But x is an integer, so the only solution is x = 6
The numbers then are : 6, 7, 8, 9, 10