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 \cdot 1 \cdot 4}{2 \cdot 1}

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

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

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

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

x = 2

The solution to the quadratic equation is:

x = 2

7 x^{2} + 10 = 0

x^{2} + 6 = - 8

p, 3 p^{2} + 9 p = - 6

x^{2} - 22 x = - 9

100 x^{2} - 49 = 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