Need help with a "completing the square" problem in quadratic factoring!

When I try to solve x^{2}+3x-10 just by factoring it from it's standard form I get x=-5 and x=2 which is good. But when I try to check that by using the "completing the square method" when you add (b/2)^{2 }to both sides I get x=5/6 which is completely different as you can see. It is said that the "completing the square method works for any problem.

Re: Need help with a "completing the square" problem in quadratic factoring!

Add $\displaystyle (b/2)^2$ to both sides of **what**? So far you don't have an equation. Please show exactly what you did. When I use "complete the square", I get the correct answer.

Re: Need help with a "completing the square" problem in quadratic factoring!

$\displaystyle x^2+3x-10=0$

$\displaystyle x^2+3x=10$

$\displaystyle x^2+3x+\frac{9}{4}=10+\frac{9}{4}$

$\displaystyle (x+\frac{3}{2})^2=\frac{49}{4}$

$\displaystyle x+\frac{3}{2}=\frac{7}{2}, \frac{-7}{2}$

$\displaystyle x=2, -5$

Re: Need help with a "completing the square" problem in quadratic factoring!

Thanks bean I forgot to square the 4