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

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

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

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

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

x = 2

Solution steps

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

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

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

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

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

x = 2

The solution to the quadratic equation is:

x = 2

x, x^{2} - x - 72 = 0

5 x + 8 = 2 x^{2}

4 x = 4 - x^{2}

(- 5 - u) (- 5 + u) = 9

x, x^{2} + 6 x - 4 = 0

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