Question

How do you solve `(x-3)^2 = 5` using the quadratic formula?

Answer 1

`(x-3)^2=5`

`x_1=3+\sqrt(5)`
`x_2=3 -\sqrt(5)`

Explanation:

`(x-3)^2=5`
`x^2-6*x+9=5`
`x^2-6*x+9-5=0`
`x^2-6*x+4=0`

`a*x^2+b*x+c=0`

`x_(1,2)=(-b +- \sqrt(b^2-4*a*c))/(2*a)`

`a=1`
`b=6`
`c=4`

`x_(1,2)=(-(-6) +- \sqrt(6^2-4*1*4))/(2*1)=`
`=(6 +- \sqrt(36-16))/(2*1)=`
`=(6 +- \sqrt(20))/(2)=`
`=(6 +- \sqrt(4*5))/(2)=`
`=(6 +- 2*\sqrt(5))/(2)=`
`=3 +-\sqrt(5)`

`x_1=3+\sqrt(5)`
`x_2=3 -\sqrt(5)`