SOLUTION: can you please check this for me? I have to solve by using the quadratic formula. 5x^2+x+10=0 a=5 b=1 c=10 b^2-4ac=1^2-4*5*10=-199 -199 is less than 0 so no solution is def

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: can you please check this for me? I have to solve by using the quadratic formula. 5x^2+x+10=0 a=5 b=1 c=10 b^2-4ac=1^2-4*5*10=-199 -199 is less than 0 so no solution is def      Log On


   

Question 81278: can you please check this for me? I have to solve by using the quadratic formula.
5x^2+x+10=0
a=5
b=1
c=10
b^2-4ac=1^2-4*5*10=-199
-199 is less than 0 so no solution is defined
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You are correct: since the discriminant is less than zero, you won't have any real solutions.

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 5x%5E2%2B1x%2B10+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A5%2A10=-199.

The discriminant -199 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -199 is + or - sqrt%28+199%29+=+14.1067359796659.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+5%2Ax%5E2%2B1%2Ax%2B10+%29