the product of two consecutive negative even integers is 80. Find the value of the lesser integer.

Let's denote the two consecutive negative even integers as x and x+2.

Since the product of these two integers is 80, we can write the equation as:

x * (x + 2) = 80

Expanding this equation gives:

x^2 + 2x = 80

Rearranging the terms to form a quadratic equation gives:

x^2 + 2x - 80 = 0

This is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 1, b = 2, and c = -80.

We can solve this equation using the quadratic formula, x = [-b ± sqrt(b^2 - 4ac)] / 2a.

Substituting the values of a, b, and c into the formula gives:

x = [-2 ± sqrt((2)^2 - 41(-80))] / 2*1 x = [-2 ± sqrt(4 + 320)] / 2 x = [-2 ± sqrt(324)] / 2 x = [-2 ± 18] / 2

This gives two possible solutions: x = 8 or x = -10.

Since we are looking for negative even integers, the lesser integer is -10.