Question

How do you solve `4+7(3x-5)^2 =1`?

Answer 1

`x_1=1/21(35-isqrt(21))`
`x_2=1/21(35+isqrt(21))`

Explanation:

Expanding after the frormula

`(a-b)^2=a^2-2ab+b^2`
we get
`63x^2-210x+148=0`
dividing all by `63`

`x^2-210/63x+148/63=0`
solving this equation

`x_1=1/21(35-isqrt(21))`

`x_2=1/21(35+isqrt(21))`

where `i^2=-1`