How do you solve #6-7x=3x²#?

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Answer 1 · 2017-06-22T08:04:29

#x=2/3, x=-3#

Put all terms onto one side of the equation so that everything is equal to #0#.
#6-7x=3x^2#
#6=3x^2+7x#
#0=3x^2+7x-6#

Factor.
#3x^2+7x-6=0#
#(3x-2)(x+3)=0#

Set each factor equal to #0#. Make sure to do this for both.
#3x-2=0#
#3x=2#
#x=2/3#

#x+3=0#
#x=-3#

Answers: #x=2/3, x=-3#

Answer 2 · 2017-06-22T14:44:19

#2/3 and - 3#

#y = 3x^2 + 7x - 6 = 0#

Use the new transforming method (Socratic Search)
Transformed equation:

#y' = x^2 + 7x - 18 = 0" "rarr (a xx c = 3(-6) = -18)#

Proceed. Find 2 real roots of y', then, divide them by #a = 3.#

Find #2# numbers knowing the sum #(-b = -7))#
and the product #(a xx c = -18)#.

They are: #2, and (-9).#

Consequently, the #2# real roots of #y# are:

#x_1 = 2/a = 2/3#, and #x_2 = -9/3 = -3#

Note. This new method avoids the lengthy factoring by grouping and solving the 2 binomials.