# Thread: quadratic formula

1. ## quadratic formula

The Quadratic Formula Explained says:

So, while factoring may not always be successful, the Quadratic Formula can always find the solution.

I'm trying to use the Quadratic Formula to solve
$1.26x^2-186.56x+34041.94=0$ but my $4ac$ is much bigger than my $b^2$ so everything under the radical evaulates to a negative number. How can I solve this?

2. Originally Posted by erinspice
The Quadratic Formula Explained says:

I'm trying to use the Quadratic Formula to solve
$1.26x^2-186.56x+34041.94=0$ but my $4ac$ is much bigger than my $b^2$ so everything under the radical evaulates to a negative number. How can I solve this?
it IS possible for a quadratic to have no real solutions. Look at x^2 + 1 = 0. This is never true for any real number x. Have you learned about complex numbers, and complex solutions? If so, the quadratic formula can still find those

3. Originally Posted by Jhevon
it IS possible for a quadratic to have no real solutions. Look at x^2 + 1 = 0. This is never true for any real number x. Have you learned about complex numbers, and complex solutions? If so, the quadratic formula can still find those
Yeah, but I need a real solution. My robot won't know what to do with a distance of xi (see this thread)! I must have made an error somewhere, because I know there has to be a real solution to my problem.

4. Originally Posted by erinspice
The Quadratic Formula Explained says:

I'm trying to use the Quadratic Formula to solve
$1.26x^2-186.56x+34041.94=0$ but my $4ac$ is much bigger than my $b^2$ so everything under the radical evaulates to a negative number. How can I solve this?
If you wish to check your answers, even those which have complex roots, download the MHF QuickSolver. The link to it is in my signature.

5. ## Complex zeros

If one sketches the quadratic function, one can see that graph of the function does not at any point intersect x-axis thus there is no real zero. However, for polynom of degree 2 there must neccesarily be two zeros. Because they are not real, they must be complex, which I guess in your case is not neccesary to calculate.