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

The solutions to the equation 2x^2-4x+4=2 are x=1

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

The zeros of 2x^2-4x+4=2 are x=1

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

x = 1

Solution steps

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

Move

2

to the left side

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

Solve with the quadratic formula

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

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

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

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

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

x = 1

The solution to the quadratic equation is:

x = 1

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

\frac{2}{3} a^{2} - 51 = - 3

n (n - 1) = 72

a, a m^{2} = a^{2} b

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

What are the solutions to the equation 2x^2-4x+4=2 ?
The solutions to the equation 2x^2-4x+4=2 are x=1
Find the zeros of 2x^2-4x+4=2
The zeros of 2x^2-4x+4=2 are x=1