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

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

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

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

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n^{2} + n - 2 = 0

n = 1, n = - 2

Solution steps

n^{2} + n - 2 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

n = \frac{- 1 - 3}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

n = 1, n = - 2

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

x, 4 x^{2} + y^{2} + 4 z^{2} - 4 y - 24 z + 36 = 0

- 3 x^{2} + 243 = 0

x, 2 x^{2} + x - 9 = 0

x^{2} - 28 = 3 x

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