Question 159567
let x = an even number.
let x+2 equal the next consecutive even number
problem states that product of 2 consecutive even numbers equals 120
your equation becomes
x * (x+2) = 120
multiplying out and it becomes
x^2 + 2*x = 120
subtracting 120 from both sides and it becomes
x^2 + 2*x - 120 = 0
since this is a quadratic equation in the form of {{{a*x^2 + b*x + c = 0}}}, you can either eyeball a solution or use the quadratic equation to solve.
the quadratic equation is {{{x = (-(b) + sqrt(b^2 - 4*a*c)) / (2*a)}}} and {{{x = (-(b) - sqrt(b^2 - 4*a*c)) / (2*a)}}}
first try to eyeball.
because the "a" in the general equation is 1, you will need factors of 
(x +/- d) * (x +/- e) = 0
since c is minus 120, these factors will be (x + d) * (x - e) or (x - d) * (x + e) since d*e must = -120.
you are looking for d and e which when added together = 2, and when multiplied together equal -120.
i didn't see it at first, but i see it now.
let d = 12 and e = -10.
when added together this makes 12 - 10 = 2
when multiplied together this makes -120.
looks like d = 12, and e = -10 are what you need.
your factors becomes (x + 12) * (x - 10) = 0
solving by multiplying these factors together you get
x^2 - 10*x + 12*x - 120 = 0
which becomes x^2 + 2*x - 120 = 0
since this is the original equation you wanted to solve, these factors look good and
x = - 12 is one of the possible answers.
x = + 10 is one of the other possible answers.
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to prove, use these values in the original equation of x*(x+2) = 120.
use x = 10 first.
the equation becomes 10 * (10 + 2) = 120 which becomes 10*12 = 120 which becomes 120 = 120 which proves the first value of 10 is good.
use x = -12 next.
the equation becomes (-12) + (-12 + 2) = 120 which becomes (-12)*(-10) = 120 which becomes 120 = 120 which proves the second value of -12 is good.
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both values are good and so your answer would be
(10 and 12), and (-12 and -10).