How do you solve #6x ^ { 2} - 30x = 36#?

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Answer 1 · 2017-03-13T19:21:33

6 and -1

#6x^2-30x-36=0#

#D=30^2-4*(-36)*6=1764=42^2#

#x_1=(30+42)/(12)=6#

#x_2=(30-42)/12=-1#

Answer 2 · 2017-03-13T20:04:37

#x = 6 " or " x =-1#

#6x^2 -30x -36 =0" "larr div 6#

#(x^2 -5x -6) =0" "larr# find the factors

#(x-6)(x+1)=0#

Either of the factors could be equal to 0.

#x-6=0" "hArr " "x=6#
#x+1 =0" "hArr" "x=-1#