How do you solve the quadratic #3x^2-2x=2x+7# using any method?

Given:#" "3x^2-2x=2x+7#

Subtract #2x# and #7# from both sides

#3x^2-4x-7=0#

7 is a prime number so does not share any common factors other than 1 with the other coefficients.

Compare to the standardised equations:

#y=ax^2+bx+c" ; "x=(-b+-sqrt(b^2-4ac))/(2a)#

Where#" "a=3"; "b=-4"; "c=-7# giving:

#x=(+4+-sqrt((-4)^2-4(3)(-7)))/(2(3))#

#x=2/3+-sqrt(16+84)/6#

#x=2/3+-10/6#

#=>x= + 7/3 and x=-1#

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Given:#" "3x^2-4x-7=0#............................Equation(1)

#color(blue)("Completing the square")#

Write as: #3(x^2-4/3x)-7=0#

Let the constant of correction be #k#

At this point #k=0#

Write as: #3(x^2-4/3x)-7+k=0#

Take the square from #x^2# outside the brackets

#3(x-4/3x)^2-7+k=0 #

Halve the coefficient of #x# so #-4/3x# becomes #-4/6x#

#"The "4/6" is not simplified on purpose"#

#3(x-4/6x)^2-7+k=0 #

Now get rid of the #x# from #-4/6x#

#color(red)(3)(xcolor(green)(-4/6))^2-7+k=0 #............................Equation(2)
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We now have to find the value of #k# so that the overall value of equation(2) is the same as that of equation(1)

The error comes from the term #color(red)(3)(color(green)(-4/6))^2# which is additional to that in the original equation(1).

Equation(1)#->3x^2-4x-7=0#
Equation(2)#-> 3x^2-4xcolor(magenta)(+(cancel(16)^4)/(cancel(12)^3))-7+k=0#
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#color(white)(.)#

So from equation(2) we have

#color(red)(3)(color(green)(-4/6))^2+k=0 " "=>" "k=-4/3larr" gets rid of the error"#

So equation(2) becomes:

#3(x-2/3)^2-7-4/3=0#

#3(x-2/3)^2-25/3=0 larr" completed square"#

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#color(blue)("Solving for "x)#

#=>(x-2/3)^2 = 25/9#

#x=2/3+-sqrt(25/9)#

#x=2/3+-5/3#

#x= 7/3 or x= -1#