Answer:
[tex]\huge\boxed{(2,\ 3);\ (-1,\ 6)}[/tex]
Step-by-step explanation:
[tex]f(x)=x^2-2x+3\to y=x^2-2x+3\\f(x)=-x+5\to y=-x+5\\\\\left\{\begin{array}{ccc}y=x^2-2x+3&(1)\\y=-x+5&(2)\end{array}\right[/tex]
substitute (1) to (2):
[tex]x^2-2x+3=-x+5[/tex] add x to both sides
[tex]x^2-2x+x+3=-x+x+5[/tex]
[tex]x^2-x+3=5[/tex] subtract 5 from both sides
[tex]x^2-x+3-5=5-5[/tex]
[tex]x^2-x-2=0\\\\x^2+x-2x-2=0\\\\x(x+1)-2(x+1)=0\\\\(x+1)(x-2)=0\iff x+1=0\ \vee\ x-2=0\\\\\boxed{x=-1\ \vee\ x=2}[/tex]
substitute the values of x to (2):
[tex]x=-1\\\\y=-(-1)+5=1+5=6\\\\\huge\boxed{(-1,\ 6)}\\\\x=2\\\\y=-2+5=3\\\\\huge\boxed{(2,\ 3)}[/tex]
