Question

How do you solve `(2x - 7) ( x + 1) = 6x - 19`?

Answer 1

`x=3/2 or 4`

Explanation:

Multiply out the Left hand side of the equation
`2x^2+2x-7x-7=2x^2-5x-7`
Therefore `2x^2-5x-7=6x-19`
Collecting like terms on the left
`2x^2-11x+12=0`
Factorise
`(2x-3)(x-4)=0`
`x=3/2 or x=4`

Answer 2

`x=4, 3/2`

Explanation:

NOTE: This is a long answer.

Solve `(2x-7)(x+1)=6x-19`

FOIL the left-hand side.

`(2x-7)(x+1)=2x*x+2x*1+(-7)*x+(-7*1)`

`(2x-7)(x+1)=2x^2+2x-7x-7`

`(2x-7)(x+1)=2x^2-5x-7`

Put the equation back together.

`2x^2-5x-7=6x-19`

Subtract `5x` from both sides.

`2x^2-6x-5x-7=-19`

`x^2-11x-7=-19`

Add `19` to both sides.

`2x^2-11x-7+19=0`

`2x^2-11x+12=0`

This is a quadratic equation, `ax+bx+c`, where `a=2`, `b=-11`, and `c=12`.

Use the quadratic formula to solve for `x`.

Quadratic Formula

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(-11)+-sqrt(-11^2-4*2*12))/(2*2)`

`x=(11+-sqrt(121-96))/4`

`x=(11+-sqrt25)/2`

`x=(11+-5)/4`

`x=16/4=4`

`x=6/4=3/2`

`x=4, 3/2`