Find all real solutions of the equation (x-5)^2-4=0 real solutions=?​

x = 7

x = 3

Step-by-step explanation:

(x - 5)² - 4 = 0

Take the (x - 5)² and foil it. By doing that, you're going to set them both against each other to multiply. The squared part means that (x - 5) is being multiplied by itself. Like so:

(x - 5) (x - 5)

Multiply the x by the x, the x by the -5, the -5 by x, and the -5 by -5.

The x times x would be x². The x times -5 would be -5x. The -5 times x would be -5x. And the -5 and -5 would be 25.

To put it in an equation: x² - 5x - 5x + 25 - 4

Simply the equation: x² - 10x + 21

Now that it's a quadratic equation, we can factor. Looking at the final term, the 21, we can find factors of 21 that add up to -10, which would be -7 and -3. They both multiply to a positive 21 and add up to a -10.

Taking this new information, we can split the equation up like so: (x -7), (x - 3)

Set both equations equal to 0.

So: x - 7 = 0

Add seven to both sides.

x = 7

x - 3 = 0

Add three to both sides

x = 3

Final x = 3, x = 7, both are real solutions