What are the solutions to the equation (2y-5)^2=0 ?

The solutions to the equation (2y-5)^2=0 are y= 5/2

Find the zeros of (2y-5)^2=0

The zeros of (2y-5)^2=0 are y= 5/2

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(2 y - 5)^{2} = 0

y = \frac{5}{2}

Decimal Notation

y = 2.5

Solution steps

(2 y - 5)^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

Solve

2 y - 5 = 0 : y = \frac{5}{2}

The solution to the quadratic equation is:

y = \frac{5}{2}

x, f (x + 4) = x^{2} - 2 x + 7

6 x (x - 1) = 40 - 5 x

x^{2} + (m - 3) x + m + 1 = 0

- 5 b^{2} + 9 b + 13 = - 6 b^{2}

- x^{2} + 5 x + 4 = 6

What are the solutions to the equation (2y-5)^2=0 ?
The solutions to the equation (2y-5)^2=0 are y= 5/2
Find the zeros of (2y-5)^2=0
The zeros of (2y-5)^2=0 are y= 5/2