Explanation
Step 1
[tex]\begin{gathered} f(x)=x^2-5x+3 \\ f(x)=-3 \end{gathered}[/tex]there is a value for y, that is the same for both equations, so
[tex]\begin{gathered} y_1=y_2 \\ x^2-5x+3=-3\Rightarrow equation\text{ (3)} \end{gathered}[/tex]solve equation (3)
[tex]\begin{gathered} x^2-5x+3=-3 \\ \text{add 3 in both sides} \\ x^2-5x+3+3=-3+3 \\ x^2-5x+6=0 \\ x^2-5x+6=0\Rightarrow ax^2+bx+c=0 \end{gathered}[/tex]hence, we can use the quadratic formula
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{replace} \\ x=\frac{+5\pm\sqrt[]{(-5)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{+5\pm\sqrt[]{25-24}}{2} \\ x=\frac{+5\pm\sqrt[]{1}}{2} \\ x=\frac{+5\pm1}{2} \\ x_1=\frac{5+1}{2}=\frac{6}{2}=3 \\ x_2=\frac{5-1}{2}=\frac{4}{2}=2 \end{gathered}[/tex]so, we have 2 values for x ( 2 and 3)
Step 2
now, find the images
a) when x= 3
[tex]\begin{gathered} f(x)=x^2-5x+3 \\ f(x)=(3)^2-5(3)+3 \\ f(x)=9-15+3 \\ f(x)=-3 \end{gathered}[/tex]b) when x= 2
[tex]\begin{gathered} f(x)=x^2-5x+3 \\ f(x)=2^2-5\cdot2+3 \\ f(x)=4-10+3 \\ f(x)=-3 \end{gathered}[/tex]therefore the solutions are
[tex]\begin{gathered} (2,-3)\text{ ;(3,-3)} \\ \end{gathered}[/tex]I hope this helps you