For this case we must solve the following quadratic equation:
[tex]x ^ 2 + 2x-6 = 0[/tex]
Where:
[tex]a = 1\\b = 2\\c = -6[/tex]
The solution is given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Substituting:
[tex]x = \frac {-2 \pm \sqrt {2 ^ 2-4 (1) (- 6)}} {2 (1)}\\x = \frac {-2 \pm \sqrt {4 + 24}} {2}\\x = \frac {-2 \pm \sqrt {28}} {2}\\x = \frac {-2 \pm \sqrt {2 ^ 2 * 7}} {2}\\x = \frac {-2 \pm2 \sqrt {7}} {2}\\x = -1 \pm \sqrt {7}[/tex]
Thus, we have two roots:
[tex]x_ {1} = - 1+ \sqrt {7}\\x_ {2} = - 1- \sqrt {7}[/tex]
ANswer:
[tex]x_ {1} = - 1+ \sqrt {7}\\x_ {2} = - 1- \sqrt {7}[/tex]
