The given system of functions is:
[tex]\begin{gathered} f(x)=2x-13 \\ g(x)=x^2-6x+3 \end{gathered}[/tex]The solution to this system is given by:
[tex]f(x)=g(x)[/tex]Replace the functions and solve for x, as follows:
[tex]\begin{gathered} 2x-13=x^2-6x+3 \\ \text{ Subtract 2x from both sides} \\ 2x-2x-13=x^2-6x-2x+3 \\ -13=x^2-8x+3 \\ \text{ Add 13 to both sides} \\ -13+13=x^2-8x+3+13 \\ 0=x^2-8x+16 \end{gathered}[/tex]Now, apply the quadratic formula to find the x-values:
[tex]\begin{gathered} \text{ The function is writen in the form} \\ ax^2+bx+c=0 \\ a=1,b=-8,c=16 \\ \text{ The quadratic function is} \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ x=\frac{-(-8)\pm\sqrt{(-8)^2-4(1)(16)}}{2(1)} \\ \\ x=\frac{8\pm\sqrt{64-64}}{2} \\ \\ x=\frac{8\pm\sqrt{0}}{2} \\ \\ x=\frac{8\pm0}{2} \\ \\ x=\frac{8}{2} \\ \\ x=4 \end{gathered}[/tex]Then, the solution is x=4.
Let's check:
[tex]\begin{gathered} f(4)=2*4-13=8-13=-5 \\ g(4)=4^2-6*4+3=16-24+3=18-24=-5 \\ f(4)=g(4)=-5 \end{gathered}[/tex]The answer is x=4.