What are the solutions to the equation r^2-4r+4=0 ?

The solutions to the equation r^2-4r+4=0 are r=2

Find the zeros of r^2-4r+4=0

The zeros of r^2-4r+4=0 are r=2

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r^{2} - 4 r + 4 = 0

r = 2

Solution steps

r^{2} - 4 r + 4 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 4 = 0

r_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 1}

r = \frac{- ( - 4 )}{2 \cdot 1}

\frac{- ( - 4 )}{2 \cdot 1} = 2

r = 2

The solution to the quadratic equation is:

r = 2

x^{5} - 2 x^{4} - x^{3} - 2 x^{2} - 20 x + 24 < 0

\frac{4 + i}{5 + i}

9 x^{2} - 9 x + 2 = 0

\frac{x + 4}{x - 2} + 22 \geq 0

\frac{3}{2} \cdot \frac{2}{3} x = 12

What are the solutions to the equation r^2-4r+4=0 ?
The solutions to the equation r^2-4r+4=0 are r=2
Find the zeros of r^2-4r+4=0
The zeros of r^2-4r+4=0 are r=2