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^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)

To solve the quadratic equation b^2 - 10b + 25 = 0, we can use the quadratic formula. The quadratic formula is given by:

b = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = 1, b = -10, and c = 25. Substituting these values into the formula, we get:

b = (-(-10) ± √((-10)^2 - 4(1)(25))) / (2(1))

Simplifying further:

b = (10 ± √(100 - 100)) / 2

b = (10 ± √0) / 2

Since √0 = 0, we have:

b = (10 ± 0) / 2

Therefore, b = 10/2 = 5.

Hence, the solution to the quadratic equation b^2 - 10b + 25 = 0 is b = 5.