SOLUTION: Hey guys, So I was on your website, practicing solving quadratics by completing the square (as you do) and I ran in to a problem. I had done this: {{{2x^2+12x+20=0}}} {{{x^2

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Hey guys, So I was on your website, practicing solving quadratics by completing the square (as you do) and I ran in to a problem. I had done this: {{{2x^2+12x+20=0}}} {{{x^2      Log On


   

Question 443130: Hey guys,
So I was on your website, practicing solving quadratics by completing the square (as you do) and I ran in to a problem. I had done this:
2x%5E2%2B12x%2B20=0
x%5E2%2B6x%2B10=0
%28x%5E2%2B6x%2B9%29%2B10-9=0 (where the nines are b/2^2)
%28x%2B3%29%5E2=-1+
What do I do now that there's a negative number on the right side? I can't square root both! HELP!!

Answer by Gogonati(855) About Me  (Show Source):
You can put this solution on YOUR website!
If you find the determinant of this quadratic equation D=6%5E2-4%2A1%2A10=-4%3C0.
We answered that this quadratic equation has no real solution.
If you know the complex numbers you can find the complex roots.
x=3%2Bi and x=3-i where i=sqrt%28-1%29.