How do you solve #6x^2=17x-12#?

1 Answer
EZ as pi
Jun 15, 2016

#(3x-4)(2x-3)#

Explanation:

As soon as you see an #x^2# term, it is a quadratic equation and it has to be made equal to 0.

#6x^2 -17x+12 = 0#

Find factors of 6 and 12 which add to 17.

We find factors of 6 and 12 and cross multiply
#3" " 4 rArr 2 xx 4 = 8#
#2" " 3 rArr 3 xx 3= 9 " " 8 + 9 = 17#

The signs in the brackets will be the same, (because of the + with the 12, they will both be -, (because of the sign with -17).

#(3x-4)(2x-3)#

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