### Instruction: Find $c,$ given that $\lfloor c\rfloor$ satisfies \[2x^2 + 17x - 55 = 0\]and $\{ c \} = c - \lfloor c \rfloor $ satisfies \[6x^2 - 23x + 7 = 0.\].### Response: This problem involves both quadratic equations and floor functions, so I will try to use factoring and/or the quadratic formula to find the roots of the equations, and then see how the floor function affects them.
First, I will look at the equation involving $\lfloor c\rfloor,$ which is $2x^2 + 17x - 55 = 0.$
I notice that this equation can be factored as $(x+11)(2x-5) = 0,$ so the roots are $x = -11$ and $x = 5/2.$
Since $\lfloor c \rfloor$ is an integer, it cannot be equal to $5/2,$ so the only possible value for $\lfloor c \rfloor$ is $-11.$
Now, I will look at the equation involving $\{ c \},$ which is $6x^2 - 23x + 7 = 0.$
This equation does not seem to be easily factorable, so I will use the quadratic formula to find the roots.
The quadratic formula says that the roots are given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\] where $a = 6, b = -23,$ and $c = 7.$
Plugging in these values, I get \[x = \frac{23 \pm \sqrt{529 - 168}}{12} = \frac{23 \pm \sqrt{361}}{12}.\]
Since $\sqrt{361} = 19,$ the roots are \[x = \frac{23 \pm 19}{12}.\]
Simplifying, I get $x = 7/2$ or $x = 1/3.$
Since $\{ c \}$ is the fractional part of $c,$ it must be between $0$ and $1,$ so the only possible value for $\{ c \}$ is $1/3.$
Therefore, $c = \lfloor c \rfloor + \{ c \} = -11 + 1/3 = -10 \frac{2}{3}.$