Question

The graph of `f\left(x\right)=2x^{2}+6x-20`is shown on the grid. Which value of `x` is a solution to `f\left(x\right)=0`?

Answer 1

To find the values of \( x \) for which \( f(x) = 0 \) in the equation \( f(x)=2x^{2}+6x-20 \), we need to solve the quadratic equation:

\[
2x^{2} + 6x - 20 = 0
\]

We can simplify this equation by dividing all terms by 2:

\[
x^{2} + 3x - 10 = 0
\]

Next, we will factor the quadratic if possible. We need to find two numbers that multiply to \(-10\) (the constant term) and add to \(3\) (the coefficient of \(x\)). Those numbers are \(5\) and \(-2\).

Thus, we can factor the quadratic as:

\[
(x + 5)(x - 2) = 0
\]

Setting each factor to zero gives us the solutions:

1. \( x + 5 = 0 \) \(\Rightarrow x = -5\)
2. \( x - 2 = 0 \) \(\Rightarrow x = 2\)

Therefore, the solutions to \( f(x) = 0 \) are \( x = -5 \) and \( x = 2 \).

If we are considering the graph of \( f(x) \), the x-intercepts of the graph (where \( f(x) = 0 \)) will correspond to these solutions.

In conclusion, the values of \( x \) that are solutions to \( f(x) = 0 \) are:

\[
\boxed{-5} \quad \text{and} \quad \boxed{2}
\]