\(f(x)=\quad 2{ x }^{ 2 }+3x+5\)
Let's called the input 'z'
When we plug 'z' in the function we get ;
\(f(z)=\quad 2{ z }^{ 2 }+3z+5\)
And we know that, this is equal to 19, so ;
\(2{ z }^{ 2 }+3z+5=\quad 19\)
Let's rearrange the equation.
\(2{ z }^{ 2 }+3z+5=\quad 19\\ \\ 2{ z }^{ 2 }+3z=\quad 19-5\\ \\ 2{ z }^{ 2 }+3z=\quad 14\\ \\ 2{ z }^{ 2 }+3z-14=\quad 0\)
So we have a quadratic equation here.
We'll use this formula to solve it :
\(\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }\)
The formula is used in equation formed like this :
\(a{ x }^{ 2 }+bx+c=0\)
In our equation,
a=2 , b=3 and c=-14
Let's plug in the values in the formula to solve,
\(a=2\quad b=3\quad c=-14\\ \\ \frac { -3\pm \sqrt { 9-(4\cdot 2\cdot -14) } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9-(-112) } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9+112 } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 121 } }{ 4 } \\ \\ \frac { -3\pm 11 }{ 4 }\)
So,
\(z=\quad \frac { -3+11 }{ 4 } \quad ,\quad \frac { -3-11 }{ 4 } \\ \\ z=\quad \frac { 8 }{ 4 } \quad ,\quad \frac { -14 }{ 4 } \\ \\ z=\quad 2,\quad -\frac { 7 }{ 2 }\)
So the input can be both, 2 and \(-\frac{7}{2}\)