Question

How do you solve `6x^2 + 2x = 104`?

Answer 1

`x=-4, -13/3`

Explanation:

Subtract `104` from both sides to get
`6x^2+2x-104=104-104` => `6x^2+2x-104=0`.

Apply the quadratic formula.

`a=6`
`b=2`
`c=-104`

`x=(-2±√2^2-4*6*(-104))/(2*6)`

Simplify to get
`x=(-2±√2^2+2496)/12`

Simplify even more to get `x=(-2±√2500)/12`
`x=(-2±50)/12`

If `x=(-2+50)/12` then `x=-4`.
If `x=(-2-50)/12` then `x=-13/3`

Yay!

Answer 2

4 and `- 13/3`

Explanation:

Use the new Transforming Method (Socratic, Google Search).
`6x^2 + 2x - 104 = 0`
`y = 3x^2 + x - 52 = 0`
Transformed equation:
`y' = x^2 + x - 156 = 0`
Proceeding. Find the 2 real roots of y', then, divide them by a = 3.
Find 2 real roots, that have opposite signs (ac < 0), knowing their sum (-b = -1) and their product (ac = - 156). They are 12 and - 13.
Back to y, the 2 real roots of y are:
`x1 = 12/a = 12/3 = 4`, and `x2 = - 13/a = - 13/3`