Question

How do you solve `x^2-6x+25=0`?

Answer 1

This has no real solution. It does have imaginary solutions however.

Explanation:

You can see this in two ways.

1) Without the discriminant
`x^2-6x+9+16=0 \color(white)(.............)`(Completing the square)
`(x-3)^2+16=0`
`(x-3)^2=-16`
`(x-3)=+-sqrt(-16)`

Since `sqrt(-16)` doesn't exist (at least not with real numbers), this equation has no real solutions.

If you want the imaginary solutions:
`(x-3)=+-sqrt(-16)`
`x-3=+-sqrt(16)*sqrt(-1)`
`x-3=+-4*i`
`x=+-4i+3`

2) With the discriminant
The discriminant is: `D = b^2-4ac` if you have a quadratic equation of the form `ax^2+bx+c=0`.
It can be proven that if this discriminant is less than 0, the equation has no solutions:
`D = (-6)^2-4*1*25=36-100=-64 < 0`