SOLUTION: solve this equation: 2x^2+3x+19=0

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Question 845343: solve this equation:
2x^2+3x+19=0
Answer by pmesler(52) About Me  (Show Source):
You can put this solution on YOUR website!
This is a simple quadratic equation. To solve this you simply use the quadratic formula
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B3x%2B19+=+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=%283%29%5E2-4%2A2%2A19=-143.

The discriminant -143 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 -143 is + or - sqrt%28+143%29+=+11.9582607431014.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B3%2Ax%2B19+%29



The roots are complex because the discriminant is negative. Anytime you have a negative square root, you have an imaginary number. A complex number is a real number plus an imaginary number of the form
a+-bi where a and b are real and i is the imaginary unit sqrt(-1).
In this case the roots are -3+11.95i/4 and -3-11.95i/4.