For this case we must find the solutions of the following quadratic equation:
[tex]x ^ 2-5x-1 = 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 = -5\\c = -1[/tex]
Substituting:
[tex]x = \frac {- (- 5) \pm \sqrt {(- 5) ^ 2-4 (1) (- 1)}} {2 (1)}\\x = \frac {5 \pm \sqrt {25 + 4}} {2}\\x = \frac {5\pm \sqrt {29}} {2}[/tex]
Finally, the roots are:
[tex]x_ {1} = \frac {5+ \sqrt {29}} {2}\\x_ {2} = \frac {5- \sqrt {29}} {2}[/tex]
Answer:
[tex]x_ {1} = \frac {5+ \sqrt {29}} {2}\\x_ {2} = \frac {5- \sqrt {29}} {2}[/tex]
