For this case we have a function of the form [tex]y = f (x).[/tex]
Where:
[tex]f (x) = 2x ^ 2 + 3x + 5[/tex]
They tell us that the function has a value of 19, and we want to know the values of the input, that is:
[tex]2x ^ 2 + 3x + 5 = 19\\2x ^ 2 + 3x + 5-19 = 0\\2x ^ 2 + 3x-14 = 0[/tex]
We apply the formula of the resolvent:
[tex]a = 2\\b = 3\\c = -14[/tex]
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}\\x = \frac {-3 \pm \sqrt {3 ^ 2-4 (2) (- 14)}} {2 (2)}\\x = \frac {-3 \pm \sqrt {9 + 112}} {4}\\x = \frac {-3 \pm \sqrt {121}} {4}\\x = \frac {-3 \pm11} {4}[/tex]
We have two roots:
[tex]x_ {1} = \frac {-3 + 11} {4} = \frac {8} {4} = 2\\x_ {2} = \frac {-3-11} {4} = \frac {-14} {4} = - \frac {7} {2}[/tex]
Answer:
The inputs of the function [tex]y = 19[/tex]are:
[tex]x_ {1} = 2\\x_ {2} = - \frac {7} {2}[/tex]
