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

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

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

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

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

n = - 2, n = 1

Solution steps

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

Refine

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 (- 1) \cdot 2 = 3

n_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 ( - 1 )}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 3}{2 ( - 1 )}, n_{2} = \frac{- ( - 1 ) - 3}{2 ( - 1 )}

n = \frac{- ( - 1 ) + 3}{2 ( - 1 )} : - 2

n = \frac{- ( - 1 ) - 3}{2 ( - 1 )} : 1

The solutions to the quadratic equation are:

n = - 2, n = 1

- 5 w^{2} - 8 w + 4

x^{2} + 18 x + 97 = 0

- 2 = - (n - 8)

- 4 (- c - 8) = - 4 c - 8

\frac{x - 4}{x ^{2} - 64} > 0

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