The integers are (n-1), n, (n+1),  and

    (n-1)*(n+1) = 4n + 59,

or

    n^2 - 1 = 4n + 59,

which implies

    n^2 - 4n - 60 = 0,

    (n-10)*(n+6) = 0.


So, the middle integer can be either 10 or -6.


And since the integers are negative, the only possibility is that the three numbers are -7, -6, -5.


This and only this triple satisfies the problem's condition.

Let the middle integer be x, then the first one is x-1 and the third one is x+1.

x-1 = smallest
x = middle
x+1 = largest

The product of the first and third of three negative consecutive integers 

That's (x-1)(x+1)

is 59 more than 4 times the second integer. Find the integers.

That's 4x + 59

(x-1)(x+1) = 4x + 59

x2 - 1 = 4x + 59
x2 - 4x - 60 = 0
(x + 6)(x - 10) = 0
x + 6 = 0;  x - 10 = 0
    x = -6;      x = 10

We are told that the integers are negative, so we discard x=10

x-1 = smallest = -6-1 = -7
x = middle     = -6  
x+1 = largest  = -6+1 = -5

Checking the words:

The product of the first and third...

(-7)(-5) = 35

...is 59 more than 4 times the second integer. 

That's 59 more than 4 times -6, which is -24.  Sure enough,'
it's true that (-24) + 59 = 35.

Edwin