Algebra - Factoring Quadratic Equations

Question

I'm in algebra and this problem was under the lesson Factoring to Solve Quadratic Equations. The problem is the following:

The product of two consecutive numbers is 14 less than 10 times the smaller number. Find each number.

I've come up with $x$ and $10x-14$. I know that there should be two variables, but that's all I could come up with as far as an expression goes. If anyone can give me a better expression or teach me how to solve this it'd be greatly appreciated. Thank you.

Answer 1

Call the smaller number $x$. Then we have

$$x(x+1) +14 = 10x$$

$$x^2 -9x + 14 = 0$$

$$(x -7)(x-2) = 0$$

so the solutions are $x=7$ or $x=2$ and so the smaller number is either $2$ or $7$. The larger number will obviously be $3$ or $8$ respectively.

Answer 2

Consecutive integers are x and x+1.

Product is $x(x+1)$

Thus the equation to solve is $x(x+1)=10x-14$