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`