x^2-4x+32=0.

Question

would the final answer be 2 +or- 2 sqrt7 ?

Answer 1

x^2-4x+32=0

a =1
b= -4
c= 32

4 +or- sq root (-4)^2 -4(1)(32)
4 +or- sq root 16-128
4 +or- sq root -112

you can't take the sq root of a (-) number so you can't solve this :/

How did you get 2 +or- 2 sqrt7.
Perhaps I went wrong somewhere.

Answer 2

hmmm. what's b^2-4ac? 16-128 = -112

So, x = 2 ± 2√(-7) = 2 ± 2√7 i

gotta watch those pesky minus signs

Answer 3

the original question was
x^2-4x=32

can you help me solve?

Answer 4

as I suspected, that makes it x^2-4x-32=0
since 32=8*4,

(x-8)(x+4) is the factorization