For this case we must solve the following equation of the second degree:
[tex]x ^ 2 + 6x-16 = 0[/tex]
The solution is given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Where:
[tex]a = 1\\b = 6\\c = -16[/tex]
Substituting:[tex]x = \frac {-6\pm \sqrt {6 ^ 2-4 (1) (- 16)}} {2 (1)}[/tex]
[tex]x = \frac{-6\pm\sqrt{36+64}}{2}[/tex]
[tex]x = \frac {-6 \pm \sqrt {100)}} {2}\\x = \frac {-6 \pm10} {2}[/tex]
So:
[tex]x_ {1} = \frac {-6 + 10} {2} = \frac {4} {2} = 2\\x_ {2} = \frac {-6-10} {2} = \frac {-16} {2} = - 8[/tex]
Answer:
[tex]x_ {1} = 2\\x_ {2} = - 8[/tex]
