For this case we have to:
Given the quadratic equation of the form:
[tex]ax ^ 2 + bx + c = 0[/tex]
The roots are given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
If we have: [tex]x ^ 2 + 14x + 17 = -96[/tex]
We can rewrite it in the following way:
[tex]x ^ 2 + 14x + 17 + 96 = 0\\x ^ 2 + 14x + 113 = 0[/tex]
Where:
[tex]a = 1\\b = 14\\c = 113[/tex]
Where we have:
[tex]x = \frac {-14 \pm \sqrt {(14) ^ 2-4 (1) (113)}} {2 (1)}[/tex]
[tex]x = \frac {-14 \pm \sqrt {(196-452)}} {2}[/tex]
[tex]x = \frac {-14 \pm \sqrt {-256}} {2}[/tex]
By definition: [tex]\sqrt {-1} = i[/tex]
[tex]x = \frac {-14 \pm \sqrt {256} i} {2}[/tex]
[tex]x = \frac {-14 \pm16i} {2}[/tex]
[tex]x = \frac {-14} {2} \pm \frac {16i} {2}[/tex]
[tex]x = -7 \pm8i[/tex]
Thus, the roots are given by imaginary numbers:
[tex]x_ {1} = - 7 + 8i\\x_ {2} = - 7-8i[/tex]
Answer:
[tex]x_ {1} = - 7 + 8i\\x_ {2} = - 7-8i[/tex]
