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October 1, 2014

October 1, 2014

Posted by **Anonymous** on Sunday, April 18, 2010 at 9:52pm.

- 12th grade algebra -
**PsyDAG**, Sunday, April 18, 2010 at 11:04pmb^2 - 10b + 25 = 0

(b -5)(b - 5) = 0

Can you work it from there?

- 12th grade -
**Anonymous**, Monday, April 26, 2010 at 12:39pmHow to solve 2nd degree equations easily:

Ax^2+Bx+c=0

x=(-B+/-sqrt(B^2-4AC))/2A

- 12th grade -
**Nolan**, Tuesday, April 27, 2010 at 5:50pmI don't agree Anonymous.

To solve 2nd degrees equations (aka find the roots of a trinomial function), you have first to put it to this form :

ax² + bx + c = 0 (where a, b and c are reals)

Then you have to calculate its discriminant (its delta) :

Δ = b² - 4ac

If Δ < 0, there is no real solution.

If Δ = 0, there is one solution :

x = -b/(2a)

If Δ > 0, there are two solutions :

x = (-b + sqrt(Δ))/(2a)

or x = (-b - sqrt(Δ))/(2a)

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