Question

How do you find the zeros, if any, of `y= 2x^2 -12x+ 23`using the quadratic formula?

Answer 1

`Delta<0`
`:. cancel(EE)x in RR`

The equation has not solutions in `RR`

or, if the excercise allow it

`x_(1,2)=3+-isqrt(10)/2 in CC`

Explanation:

Given

`y=ax^2+bx+c=2x^2-12x+23=0`

The Quadratic Formula tells:

`x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)=(-b+-sqrt(Delta))/(2a)`

where:

`a=2`
`b=-12`
`c=23`

`:. x_(1,2)=(12+-sqrt(144-4*2*23))/4=(12+-sqrt(144-4*2*23))/4=(12+-sqrt(144-184))/4=(12+-sqrt(-40))/4`

Now, `Delta<0` therefore `cancel(EE)x in RR`

You have solutions in `CC`, if the exercise allows them

`x_(1,2)=(12+-sqrt(-1)sqrt(40))/4=(12+-isqrt(4*10))/4=`
`=(cancel(12)^6+-icancel(2)sqrt(10))/cancel(4)^2=3+-isqrt(10)/2`