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

The solutions to the equation n^2-2n-4=0 are n=1+sqrt(5),n=1-sqrt(5)

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

The zeros of n^2-2n-4=0 are n=1+sqrt(5),n=1-sqrt(5)

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

n = 1 + 5, n = 1 - 5

Decimal Notation

n = 3.23606 \dots, n = - 1.23606 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 4) = 25

n_{1, 2} = \frac{- ( - 2 ) \pm 2 5}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 2 ) + 2 5}{2 \cdot 1}, n_{2} = \frac{- ( - 2 ) - 2 5}{2 \cdot 1}

n = \frac{- ( - 2 ) + 2 5}{2 \cdot 1} : 1 + 5

n = \frac{- ( - 2 ) - 2 5}{2 \cdot 1} : 1 - 5

The solutions to the quadratic equation are:

n = 1 + 5, n = 1 - 5

2 (6 x + 2) = 25 (x - 6)^{2}

w^{2} - 6 w + 8 = 0

40 = 50 + 2 t - 0.0625 t^{2}

4 x + 8 x^{2} + 2 = - 3 x

x, 3 x^{2} + 4 x y + y^{2} = 6

What are the solutions to the equation n^2-2n-4=0 ?
The solutions to the equation n^2-2n-4=0 are n=1+sqrt(5),n=1-sqrt(5)
Find the zeros of n^2-2n-4=0
The zeros of n^2-2n-4=0 are n=1+sqrt(5),n=1-sqrt(5)