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Algebra 15 Online
OpenStudy (anonymous):

how do i find the possible values of x in 6x^2 + 432 = 0

OpenStudy (anonymous):

the solutions will not be real numbers they will be complex numbers start with \[6x^2=-432\] then divide both sides by \(6\)

OpenStudy (anonymous):

you get \[x^2=-72\] so the solutions are complex

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