Question

How do you solve `5x^2-7=0`?

Answer 1

`x=+-sqrt(35)/5`

Explanation:

You can solve this by isolating `x`.

`5x^2−7=0`
`5x^2−7+7=0+7`
`5x^2=7`
`1/5(5x^2)=1/5(7)`
`x^2=7/5`
`x=+-sqrt(7/5)`

You can have that as your final answer or you can rationalize it:
`x=+-sqrt(7/5)`
`x=+-sqrt(7/5*5/5)`
`x=+-sqrt((7*5)/5^2)`
`x=+-sqrt(35)/5`