Question

How do you solve using the quadratic formula `x^2-4x+10=0`?

Answer 1

Given a quadratic equation of the form
`ax^2+bx+c=0`
the quadratic roots can be evaluated using the formula
`x=(-b+-sqrt(b^2-4ac))/(2a)`

For `x^2-4x+10 = 0`

`x= (4+-sqrt(16-40))/2`

`x= 2+-sqrt(-6)`

This equation has no Real solutions
but if we are allowed Complex solutions:
`x=2+-sqrt(6)i`

Answer 2

You notice the 1st degree coefficient is even, so you can use the direct formula
`P(x)=ax^2+bx+c, β=b2`
`x1,x2={−β±sqrt(β^2−ac)}/a`
so `x_1,x_2=2±sqrt(4−10)=2±sqrt(−6)`
So you know there are no solutions in ℝ, and you have the complex conjugated solutions `z=2+sqrt(6)i` and`barz=2−sqrt(6)i`