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

2. Originally Posted by erinspice

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