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

no

angela210793 · Answer

this must be in the maths group

anonymous · Answer

There are solutions but they are imaginary. But as others said. Math group is -> that way.

waheguru · Answer

wrong group

anonymous · Answer

try goin to the math group and asking :)

anonymous · Answer

we're the writing geeks, not the math nerds, hahaha :) better ask the math group...