Question

Please help with 1 Math question!!!

Answer 1

For this case we must solve the following quadratic equation:

$x ^ 2-5x-8 = 0$

We use the quadratic formula to find the solutions:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

Where:

$a = 1\\b = -5\\c = -8$

Substituting we have:

$x = \frac {- (- 5) \pm \sqrt {(- 5) ^ 2-4 (1) (- 8)}} {2 (1)}\\x = \frac {5 \pm \sqrt {25 + 32}} {2}\\x = \frac {5 \pm \sqrt {57}} {2}$

In this way we have two roots:

$x_ {1} = \frac {5+ \sqrt {57}} {2}\\x_ {2} = \frac {5- \sqrt {57}} {2}$

Answer:

$x_ {1} = \frac {5+ \sqrt {57}} {2}\\x_ {2} = \frac {5- \sqrt {57}} {2}$