Question

Hey! If you can help here it is!

Answer 1

`x=3,x=-1`

Explanation:

Start with:

`x(x-1)-3(x-3)+3(x-2)=x+6`

Distribute the values on the outside of each of the parentheses.

`x^2-x-3x+9+3x-6=x+6`

Combine like terms.

`x^2-x+3=x+6`

Subtract
`6` from both sides of the equation.

`x^2-x-3=x`

Subtract
`x` from both sides of the equation.

`x^2-2x-3=0`

Now, we need to use our quadratic equation formula:

`ax^2+bx+c=0`

`x= (-b+/-√(b^2-4ac))/(2a)`

Identify your values.

`a=1\\b=-2\\c=-3`

Substitute.

(Solve positive)

`x= (-(-2)+√((-2)^2-4(1)(-3)))/(2(1))`

Solve.

`x= (2+√((16))/(2)`

Find the square root of
`16`.

`x=(2+4)/(2)`

Add.

`x=(6)/(2)`

Simplify.

`x=3`

~

(Solve negative)

`x= (-(-2)-√((-2)^2-4(1)(-3)))/(2(1))`

Solve.

`x= (2-√((16))/(2)`

Find the square root of
`16`.

`x=(2-4)/(2)`

Subtract.

`x=(-2)/(2)`

Simplify.

`x=-1`