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

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

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

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

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

r = \frac{5}{2}

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

Solve

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

The solution to the quadratic equation is:

r = \frac{5}{2}

21 k^{2} + 4 k = - 3 + 6 k + 9 k^{2}

3 a^{2} - 27 = 0

x^{2} - 8 x = 9 x - 12

4 x - 10 (- x - 1) = 6 (x - 2) 6 (x - 2)

x, x y^{2} + x^{2} + 4 y - 4 = 0

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