:confused: what r the integers

Not too bad. Any number multiplied by 2 will be even. 1*2, 2*2, 3*2 etc.
Consecutive even integers differs by 2 and defined by a*2 and (a+1)*2

So (a*2)*(a+1)*2 = 224 [fixed - courtesy of ISneezeFunny, thanks]

I1= a*2
I2= (a+1)*2

Just a dumb check: suppose a =3; I1=6, I2=8. They are consecutive even integers.

Solve for a and then I1 and I2

I agree. A little difficult, but you can't have 1 equation and 2 unknowns.

To go along what keepitsimplestupid said, although (a*2) + (a+1)*2 is wrong, as that's addition and the question asks for product.

2a is the first integer
2a + 2 is the next even integer.

(2a)(2a + 2) = 224

4a^2 + 4a = 224

a^2 + a = 56

a^2 + a - 56 = 0

(a + 8)(a - 7) = 0

a = -8 or 7

try -8

first even integer = 2a = -16, second even integer is -14

(-16)(-14) = 224, check.

try 7

first even integer = 2a = 14, second even integer is 16

(14)(16) = 224, check

It's really much simpler than all that. Just take the square root of 224, and the answer will be the two even numbers that "bracket" the square root.

Eaines approach is the easiest way to solve this problem.

That's the way I solved it.

I think the best way to do this is that you assign the smaller even number as X and the greater one as X + 2 since they are consecutive .
therefore:

x(x+2) = 224
and you can solve this quadratically and you'll get the answer simply
x = 14 , x = 16
so the two integers are 14 and 16