Question 121·Easy·Nonlinear Equations in One Variable; Systems in Two Variables

Subtract $14$ from both sides: $x^2+5x-14=0$. Now a solution is a value of $x$ that makes the expression on the left equal to zero.

Type $y=x^2+5x-14$ and $y=0$ in Desmos. Click both crossings: Desmos shows $(-7,0)$ and $(2,0)$. Each crossing's $x$-coordinate solves the equation. Only $2$ appears among the choices, so $2$ is a solution to the given equation. Choice A.

In $x^2+5x-14$, look for two numbers that multiply to $-14$ and add to $5$. They are $7$ and $-2$, so factor the zero equation from above:

The zero-product property says that if a product is $0$, at least one factor is $0$. Set each factor to $0$: $x+7=0\quad\text{or}\quad x-2=0$. Solve each equation: $x=-7\quad\text{or}\quad x=2$. Don't copy the sign from a factor: $x-2=0$ gives $x=2$. That is the listed solution. Choice A.