If $(x+2)(x-3)=14$, find the sum of the possible values of $x$.
Expanding the left side of the given equation, we have $x^2-x-6=14 \Rightarrow x^2-x-20=0$. Since in a quadratic  with equation of the form $ax^2+bx+c=0$ the sum of the roots is $-b/a$, the sum of the roots of the given equation is $1/1=\boxed{1}$.
1
Level 3
Algebra
Problem: If $(x+2)(x-3)=14$, find the sum of the possible values of $x$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation $(x+2)(x-3)=14$, we first expand the left side. This gives us $x^2 - 3x + 2x - 6 = x^2 - x - 6$. 

Next, we set the equation equal to 14: 

$x^2 - x - 6 = 14$. 

We then move 14 to the left side to set the equation to zero: 

$x^2 - x - 20 = 0$. 

Now, we can factor the quadratic equation. We look for two numbers that multiply to -20 and add to -1. The numbers -5 and 4 satisfy these conditions. Thus, we can factor the equation as: 

$(x - 5)(x + 4) = 0$. 

Setting each factor to zero gives us the possible values of $x$: 

$x - 5 = 0$ leads to $x = 5$, and $x + 4 = 0$ leads to $x = -4$. 

Now, we find the sum of the possible values of $x$: 

$5 + (-4) = 1$. 

Thus, the final answer is 1.<|END_OF_LATENT|> The final answer is $1$.