Find two consecutive integers such that their product is 11 more than 2 times their sum

I strongly recommend starting out by choosing an appropriate variable.

We could let the first integer be x and the next consecutive integer be x+1.

The sum of these two consec. integers is x + (x+1), or 2x+1. Twice this sum is 4x+2.

The product of these two consec. int. is x (x+1), which is 11 more than twice their sum. Writing this out symbolically,

x (x+1) = 2[2x+1] + 11

Expanding: x^2 + x = 11 + 4x + 2

x^2 + x - 4x - 13 = 0

x^2 - 3x - 13 = 0

Unfortunately, this quadratic equation, while solvable, has neither integer nor real roots. Thus, there is no solution which consists of the integer x and the next consecute integer x+1. Are you positive you have copied down this problem correctly?