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I don't know where to start I have tried using five to find the image but it doesn't give me the correct answer

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    I is not clear what your question is. – 2017-01-27
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    Well, to have five as their image means to find $\{x| x^2 + 4x = 5\}$. So find all the solutions to $x^2 + 4x = 5$. – 2017-01-27

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Definition: An image is the subset of a function's codomain which is the output of the function from a subset of its domain.

If $f : X → Y$ is a function from the set $X$ to the set $Y$, and if $x \in X$, then $f(x) = y$ is the image of $x$ under $f$.

So in our case, $x^2+4x =5 \Rightarrow x^2+4x-5=0 \Rightarrow (x+5)(x-1)=0$. Can you take it from here? Hope it helps.

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    So it would be x=-5 and x=1 – 2017-01-27
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    @MahmudBugaje Yes you are correct. – 2017-01-27
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I think you want to find the values of $x$ where $G(x) = x^2 + 4x$ equals 5. One can easily see that $x=1$ is a solution; the other one is $x=-5$. In any case, for quadratic equations you can't solve immediately, you can use the ABC-formula.

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$G(x) = 5 \implies x^2 + 4x = 5 \implies x^2 + 4x - 5 = 0$. By quadratic rule we have the values of $x = 1 , -5$.