Question

How do you solve `2(x-3)^2=8`?

Answer 1

`x=5` or `x=1`

See below

Explanation:

`2(x−3)^2=8`

Divide both sides by 2: `(x-3)^2=4`

Take the square root of each side:

`(x-3)=+-2`

`x-3=+2" "rarrx=5`
or
`x-3=-2" "rarrx=1`

`x=5` or `x=1`

Answer 2

The answer is x= 5 and 1

Explanation:

First you start by evualting the `2(x-3)^2` part of the equation.
`2(x-3)^2`=8
`2(x^2-6x+9)=8`
`2x^2-12x+18=8`
Then you move over the constant on the right side to the left .
`2x^2-12x+18-8=0`
`2x^2-12x+10=0`
Then divide the whole equation by 2.
`(2x^2-12x+10)/2=0/2`
`x^2-6x+5=0`
Then factor the equation normally.
`(x-5)(x-1)=0`