12th grade
posted by Anonymous .
Solve this quadratic equation b2  10b plus 25 equals 0

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^24AC))/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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