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

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

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

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

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

v = 2

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

v = 2

The solution to the quadratic equation is:

v = 2

3 \cdot 1 0^{- 2}

\frac{1}{3} \cdot \frac{3}{2}

2 x^{2} = 5 x + 12

5 (- 3 + 3 x) = 4 x - 26

\frac{t}{y - t} - \frac{y}{y + t}

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