In the quadratic equation #x^2+4ex=3f#, #e# and #f# are constants. What are the solutions for #x#?

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Answer 1 · 2017-09-30T06:03:39

#=-2e+-sqrt(4e^2+3f)#

#x^2+4ex=3f#

=#x^2+4ex-3f=0#

#=(-4e+-sqrt((4e)^2-4*1*(-3f)))/2#

#=(-4e+-sqrt(16e^2+12f))/2#

#=(-4e+-2sqrt(4e^2+3f))/2#

#=-2e+-sqrt(4e^2+3f)#