For this case we have the following quadratic equation:
[tex]p = -5x ^ 2 + 70x-165[/tex]
To find the roots, we equal 0.
[tex]-5x ^ 2 + 70x-165 = 0[/tex]
Where:
[tex]a = -5\\b = 70\\c = -165[/tex]
The solution will be given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Substituting we have:
[tex]x = \frac {-70 \pm \sqrt {70 ^ 2-4 (-5) (- 165)}} {2 (-5)}\\x = \frac {-70 \pm \sqrt {4900-3300}} {- 10}\\x = \frac {-70 \pm \sqrt {1600}} {- 10}\\x = \frac {-70 \pm40} {- 10}\\x = - (-7\pm4)[/tex]
We have two roots:
[tex]x_ {1} = 7-4 = 3\\x_ {2} = 7 + 4 = 11[/tex]
Answer:
[tex]x_ {1} = 3\\x_ {2} = 11[/tex]
