For this case we must solve the following quadratic equation:
[tex]x ^ 2 + 15x-36 = 0[/tex]
The equation cannot be factored.
We find the roots by the following formula:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Where:
[tex]a = 1\\b = 15\\c = -36[/tex]
Substituting the values:
[tex]x = \frac {-15 \pm \sqrt {15 ^ 2-4 (1) (- 36)}} {2 (1)}\\x = \frac {-15 \pm \sqrt {225 + 144}} {2}\\x = \frac {-15 \pm \sqrt {369}} {2}\\x = \frac {-15 \pm \sqrt {3 ^ 2 * 41}} {2}\\x = \frac {-15 \pm3 \sqrt {41}} {2}[/tex]
We have two roots:
[tex]x_ {1} = \frac {-15 + 3 \sqrt {41}} {2}\\x_ {2} = \frac {-15-3 \sqrt {41}} {2}[/tex]
Answer:
[tex]x_ {1} = \frac {-15 + 3 \sqrt {41}} {2}\\x_ {2} = \frac {-15-3 \sqrt {41}} {2}[/tex]
