How do you solve #2/3x^2+30=0#?

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Answer 1 · 2016-12-28T14:54:32

#x=+-3sqrt(5)color(white)(.)i#

Subtract 30 from both sides

#2/3x^2=-30#

Multiply both sides by #3/2#

#x^2=-45#

square root both sides

#x=+-sqrt(-45)#

But #45 = 5xx9 = 5xx3^2# giving

#x=+-3sqrt(-5)#

#x=+-3sqrt(5xx-1)#

But #sqrt(-1)=i#

#x=+-3sqrt(5)color(white)(.)i#