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

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

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

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

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v, 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

- 1 x^{2} - 2 x + 3 = 0

r, A = \frac{1}{2} r^{2} θ

ab x^{2} - a^{2} x - b^{2} x + ab = 0

0.042 x^{2} + x - 18 = 0

2 = 3 x^{2} + 1

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