How do you solve #-2x^2-80=0#?

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Answer 1 · 2016-11-08T19:35:16

#x=+-2sqrt(10)i#

#-2x^2-80=0#

#rarr x^2+40=0#

#rarr x^2=-40= i^2 * 2^2 * 10 #

#rarr x = +- i2sqrt(10) or +-2sqrt(10)i#

Answer 2 · 2016-11-08T19:40:11

#-2x^2 -80 =0#

Usually you would want a quadratic equation to be equal to 0, but in this case there is no #x# term, so we will solve it by another method.

Re-arrange to give:

#-2x^2 = 80" "larr div -2#

#x^2 = -40#

#x = +-sqrt(-40)#

There are no real solutions to this equation.