If the radical is negative, is there a solution to the quadratic equation?

Using the quadratic formula, a=1, b= -4, and c= 5.

After substituting those values into the formula, we have:
-4 ± √4^2 - 4(1)(5) / 2(1)

Simplifying the radical we get 16-20, which results in -4.
So, -4 ±√-4 / 2.

Since the radical is -4, is there a solution? If not, is there any need to go further with this problem?

Question

If the radical is negative, is there a solution to the quadratic equation?

Using the quadratic formula, a=1, b= -4, and c= 5.

After substituting those values into the formula, we have:
-4 ± √4^2 - 4(1)(5) / 2(1)

Simplifying the radical we get 16-20, which results in -4. 
So, -4 ±√-4 / 2.

Since the radical is -4, is there a solution?  If not, is there any need to go further with this problem?

anonymous · Answer

This is the wrong group! Post it in math!