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

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

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

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

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

n = 1

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

n = 1

The solution to the quadratic equation is:

n = 1

6 x^{2} + 11 x = 8

5 a^{2} - 45 = 0

3 x + 1 = x^{2} - 6 x + 9

- x^{2} + 10 x = 24

x^{2} - 3 x + 2 = 6

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