Give exact approximate solutions? HUH? to three decimal places.

y^2-20y+100=64

In terms of i and use radicals if needed.

:-) Thanks

bring the 64 over so you have

y^2 - 20y + 36

im not sure if you can factor this so just use quadratic formula

[ 20 +/- sqrt(400 -144) ]/2

= 10 +/- .5 sqrt 256

= 10 +/- 8

= 2 or 18

To solve the equation y^2 - 20y + 100 = 64, we can rearrange it by subtracting 64 from both sides:

y^2 - 20y + 36 = 0

Now let's solve the quadratic equation using the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = -20, and c = 36. Substituting these values into the formula:

y = (-(-20) ± √((-20)^2 - 4(1)(36))) / (2(1))

Simplifying further:

y = (20 ± √(400 - 144)) / 2
y = (20 ± √(256)) / 2
y = (20 ± 16) / 2

Now we have two possible solutions:

1. y = (20 + 16) / 2 = 36 / 2 = 18
2. y = (20 - 16) / 2 = 4 / 2 = 2

Therefore, the two approximate solutions for y are 18 and 2.