For this case we must solve the following quadratic equation:
[tex]x ^ 2 + 2x + 1 = 17[/tex]
Subtracting 17 from both sides of the equation we have:
[tex]x ^ 2 + 2x + 1-17 = 0\\x ^ 2 + 2x-16 = 0[/tex]
The solution of the equation is given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Where:
[tex]a = 1\\b = 2\\c = -16[/tex]
Substituting the values we have:
[tex]x = \frac {-2 \pm \sqrt {2 ^ 2-4 (1) (- 16)}} {2 (1)}\\x = \frac {-2 \pm \sqrt {4 + 64}} {2}\\x = \frac {-2 \pm \sqrt {68}} {2}\\x = \frac {-2 \pm \sqrt {2 ^ 2 * 17}} {2}\\x = \frac {-2 \pm2 \sqrt {17}} {2}\\x = -1 \pm \sqrt {17}[/tex]
Thus, we have two roots:
[tex]x_ {1} = - 1+ \sqrt {17}\\x_ {2} = - 1- \sqrt {17}[/tex]
Answer:
[tex]x_ {1} = - 1+ \sqrt {17}\\x_ {2} = - 1- \sqrt {17}[/tex]
