For this case we have the following system of equations:
[tex]y = x ^ 2-4x + 3\\y = -x + 3[/tex]
Matching we have:
[tex]x ^ 2-4x + 3 = -x + 3\\x ^ 2-4x + x + 3-3 = 0\\x ^ 2-3x = 0[/tex]
We solve by means of
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Where:
[tex]a = 1\\b = -3\\c = 0[/tex]
Substituting:
[tex]x = \frac {- (- 3) \pm \sqrt {(- 3) ^ 2-4 (1) (0)}} {2 (1)}\\x = \frac {3 \pm \sqrt {9}} {2}\\x = \frac {3 \pm3} {2}[/tex]
Finally, the roots are:
[tex]x_ {1} = \frac {3-3} {2} = 0\\x_ {2} = \frac {3 + 3} {2} = \frac {6} {2} = 3[/tex]
Answer:
[tex]x_ {1} = 0\\x_ {2} = 3[/tex]
