Question

How do you solve `7x^2 +28=53x`?

Answer 1

Factor by grouping to find zeros:

`x = 4/7` and `x=7`

Explanation:

For subtract `53x` from both sides to get:

`7x^2-53x+28 = 0`

We can split the middle term and factor by grouping as follows:

`0 =7x^2-53x+28`

`=7x^2-49x-4x+28`

`=(7x^2-49x)-(4x-28)`

`=7x(x-7)-4(x-7)`

`=(7x-4)(x-7)`

Hence zeros `x=4/7` and `x=7`

Answer 2

(4/7) and 7.

Explanation:

I use the new Transforming Method (Google, Yahoo Search).
`y = 7x^2 - 53x + 28 = 0`
Transformed equation: `y' = x^2 - 53x + 196`
Factor pairs of (ac = 196) --> (2, 98)(4, 49). This sum is 53 = -b. Then the 2 real roots of y' are: 4 and 49.
Back to original equation y, the 2 real roots are: `(4/7)` and
`(49/7 = 7)`

NOTE . There is no need to factor by grouping and solving the 2 binomials.