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

The solutions to the equation 2x^2-4x+2=0 are x=1

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

The zeros of 2x^2-4x+2=0 are x=1

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

x = 1

Solution steps

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

4 x^{2} - 5 x + 10 = 0

\frac{2 ^{5}}{3}

- \frac{113}{( 0 + 6 ) ^{2}}

- 4 (5 x + 2) + 2 x + 4 = 50

\frac{2 3 + 5 2}{3 3 - 2 2}

What are the solutions to the equation 2x^2-4x+2=0 ?
The solutions to the equation 2x^2-4x+2=0 are x=1
Find the zeros of 2x^2-4x+2=0
The zeros of 2x^2-4x+2=0 are x=1