Posted by Anonymous on Sunday, April 18, 2010 at 9:52pm.
b^2 - 10b + 25 = 0
(b -5)(b - 5) = 0
Can you work it from there?
How to solve 2nd degree equations easily:
Ax^2+Bx+c=0
x=(-B+/-sqrt(B^2-4AC))/2A
I 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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