which two numbers multiply to give you 100 and add to give you 10:

x*y = 100

x+y = 10

x = 10 - y

x(10-x) = 100

10x - x^2 = 100

. I don t remember the delta statment :)

The correct answer is:

There are no real numbers that fit these criteria.

Explanation:

Let x and y represent the two numbers we are looking for.

x*y = 100

x+y = 10

Solving for y,

x+y-x = 10-x

y = 10-x

Substituting this into our first equation, we have:

x(10-x) = 100

Using the distributive property,

x*10 - x*x = 100

10x - x² = 100

Writing this in standard form, we need to subtract 100 from each side:

10x - x² - 100 = 100 - 100

10x - x² - 100 = 0

In standard form, this is

-x²+10x-100 = 0.

This is in the form ax²+bx+c; in our equation, a = -1, b = 10 and c = -100.

We will use the quadratic formula for this:

\( x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\=\frac{-10\pm \sqrt{10^2-4(-1)(-100)}}{2(-1)} \\ \\=\frac{-10\pm \sqrt{100-400}}{-2} \\ \\=\frac{-10\pm \sqrt{-300}}{-2} \)

There is no real square root of -300, so there are no real number solutions to this.

x*y = 100

x + y = 10

y = 10 - x

x(10 - x) = 100

-x^2 + 10 - 100 = 0

delta = 10^2 - 4(-1)(-100) = 100 + 4(-100) = 100 - 400 = -300

Which means if i haven ’ t made a mistake: