WHAT 2 NUMBERS EQUALS -272 WHEN YOU MULTIPLY THEM AND THOSE SAME 2 NUMBERS EQUAL 2 WHEN YOU ADD THEM?

Divide -272 by 2.

Then check your work.
THEN, make sure that when you add them, they equal two. If not, then you either did the math wrong, or my method is wrong. Either way, good luck.

Use some intuition on these kind

You probably want integers.
You know one is positive , the other negative.
Since their sum is 2 they must be "almost" opposites, and they must be only 2 apart in their absolute value.

take √272 = 16.49...
so try 16 and 17

try the numbers +17 and -16

(+17)(-16) = -272
but +17 + (-16) = 1 , not 2

so there are no integers.

so let's try some algebra
x+y = 2 , -----> y = 2-x
xy = -272

x(2-x) = -272
x^2 - 2x - 272 = 0

x = 1 ± √273

so try x = 1 + √273 = 17.5227
then y = -17.5227

(17.5227)(-15.5227) = -271.9996
(17.5227) + (-15.5227) = 2

so the two numbers are 1+√273 and 2-√273

Let's solve this math problem step by step:

Let's assume the two numbers are 'x' and 'y'.

According to the given information:
1) When you multiply the two numbers, the result is -272:
xy = -272 -- Equation 1

2) When you add the two numbers, the result is 2:
x + y = 2 -- Equation 2

To find the values of 'x' and 'y', we can use a method called substitution or elimination.

Method 1: Substitution
From Equation 2, let's rearrange it to solve for 'x' or 'y':
x = 2 - y

Now replace 'x' in Equation 1 with the value obtained for 'x':
(2 - y)y = -272

Simplify this equation:
2y - y^2 = -272

Rearrange it to form a quadratic equation:
y^2 - 2y - 272 = 0

Now we can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. However, I will use the quadratic formula to find the value of 'y':

The quadratic formula is given by:
y = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = -2, and c = -272.

Substitute these values into the quadratic formula:
y = (-(-2) ± √((-2)^2 - 4(1)(-272))) / (2(1))

Simplify further:
y = (2 ± √(4 + 1088)) / 2
y = (2 ± √1092) / 2

Now let's simplify the square root:
y = (2 ± √(4 * 273)) / 2
y = (2 ± 2√273) / 2

Divide both numerator and denominator by 2:
y = 1 ± √273

So we have two possible values for 'y': 1 + √273 and 1 - √273.

Now substitute these values back into Equation 2 to find the corresponding values of 'x':
For y = 1 + √273:
x + (1 + √273) = 2
x + √273 = 1
x = 1 - √273

For y = 1 - √273:
x + (1 - √273) = 2
x - √273 = 1
x = 1 + √273

Therefore, the two numbers that satisfy the given conditions are:
1) x = 1 - √273, y = 1 + √273
2) x = 1 + √273, y = 1 - √273