Respuesta :

[tex]f(x)=\quad 2{ x }^{ 2 }+3x+5[/tex]

Let's called the input 'z'

When we plug 'z' in the function we get ;

[tex]f(z)=\quad 2{ z }^{ 2 }+3z+5[/tex]

And we know that, this is equal to 19, so ;

[tex]2{ z }^{ 2 }+3z+5=\quad 19[/tex]

Let's rearrange the equation.

[tex]2{ z }^{ 2 }+3z+5=\quad 19\\ \\ 2{ z }^{ 2 }+3z=\quad 19-5\\ \\ 2{ z }^{ 2 }+3z=\quad 14\\ \\ 2{ z }^{ 2 }+3z-14=\quad 0[/tex]

So we have a quadratic equation here.

We'll use this formula to solve it :

[tex]\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } [/tex]

The formula is used in equation formed like this :

[tex]a{ x }^{ 2 }+bx+c=0[/tex]

In our equation,

a=2 , b=3 and c=-14

Let's plug in the values in the formula to solve,

[tex]a=2\quad b=3\quad c=-14\\ \\ \frac { -3\pm \sqrt { 9-(4\cdot 2\cdot -14) } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9-(-112) } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9+112 } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 121 } }{ 4 } \\ \\ \frac { -3\pm 11 }{ 4 } [/tex]

So,

[tex]z=\quad \frac { -3+11 }{ 4 } \quad ,\quad \frac { -3-11 }{ 4 } \\ \\ z=\quad \frac { 8 }{ 4 } \quad ,\quad \frac { -14 }{ 4 } \\ \\ z=\quad 2,\quad -\frac { 7 }{ 2 } [/tex]

So the input can be both, 2 and [tex] -\frac{7}{2} [/tex]

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