For this case we have the following system of equations:
[tex]y = x ^ 2-3\\y = -2x + 4[/tex]
Matching we have:
[tex]x ^ 2-3 = -2x + 4\\x ^ 2 + 2x-3-4 = 0\\x ^ 2 + 2x-7 = 0[/tex]
Where:
[tex]a = 1\\b = 2\\c = -7[/tex]
The solution is given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}[/tex]
Substituting:
[tex]x = \frac {-2 \pm \sqrt {(2) ^ 2-4 (1) (- 7)}} {2 (1)}\\x = \frac {-2 \pm \sqrt {4 + 28}} {2}\\x = \frac {-2 \pm \sqrt {32}} {2}\\x = \frac {-2 \pm \sqrt {4 ^ 2 * 2}} {2}\\x = \frac {-2 \pm4 \sqrt {2}} {2}\\x = -1 \pm2 \sqrt {2}[/tex]
We have two roots:
[tex]x_ {1} = - 1 + 2 \sqrt {2}\\x_ {2} = - 1-2 \sqrt {2}[/tex]
We find the values of "y":
[tex]y_ {1} = - 2 (-1 + 2 \sqrt {2}) + 4 = 2-4 \sqrt {2} + 4 = 6-4 \sqrt {2}\\y_ {2} = - 2 (-1-2 \sqrt {2}) + 4 = 2 + 4 \sqrt {2} + 4 = 6 + 4 \sqrt {2}[/tex]
The system solution is:
[tex](-1 + 2 \sqrt {2}; 6-4 \sqrt {2})\\(-1-2 \sqrt {2}; 6 + 4 \sqrt {2})[/tex]
Answer:
[tex](-1 + 2 \sqrt {2}; 6-4 \sqrt {2})\\(-1-2 \sqrt {2}; 6 + 4 \sqrt {2})[/tex]
