How do you solve #(-2x+5)^2 = -8#?

1 Answer

Answer:

Refer to explanation

Explanation:

Basically this is impossible in the set of real numbers because the first part is always positive and the second part is always negative.

Now in the set of complex numbers we have that

#(-2x+5)^2=-8=>(-2x+5)^2=8i^2=> -2x+5=+-sqrt(8i^2)=>2x=5+-isqrt8=>x=1/2(5+-isqrt8)#

where #i# is the imaginary unit with #i^2=-1#