Question

Question is in the picture. Me have no clue. Please help.

Answer 1

I can’t see all the options in the Dropbox but think of

PEMDAS

Answer 2

For this case we have the following quadratic equation
`0 = x ^ 2-10x-27`, which can be written like:
`x ^ 2-10x-27 = 0`, thus, it is of the form:

`ax ^ 2 + bx + c = 0`

Where:

`a = 1\\b = -10\\c = -27`

The roots will be:

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

Substituting we have:

`x = \frac {- (- 10) \pm \sqrt {(- 10) ^ 2-4 (1) (- 27)}} {2 (1)}\\x = \frac {10 \pm \sqrt {100-4 (-27)}} {2}\\x = \frac {10 \pm \sqrt {100 + 108}} {2}\\x = \frac {10 \pm \sqrt {208}} {2}\\x = \frac {10 \pm \sqrt {16 * 13}} {2}\\x = \frac {10 \pm4 \sqrt {13}} {2}\\x = 5 \pm 2 \sqrt {13}`

Thus, we have two solutions given by:

`x_ {1} = 5 + 2 \sqrt {13} = - 2.2111\\x_ {2} = 5-2 \sqrt {13} = 12.2111`

Answer:

The negative solution is between -3 and -2

The positive solution is between 12 and 13