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:
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)