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

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

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

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

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

t = 2

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

t = 2

The solution to the quadratic equation is:

t = 2

(\frac{8}{5})^{- 1}

lo g_{2} (9 4^{8})

2 x^{2} - 12 x + 32 = x^{2}

4 - (- 2)^{2}

\frac{1}{12 m ^{5} n ^{3}} + \frac{6}{18 mn}

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