What are the solutions to the equation n^2-2n-28=0 ?

The solutions to the equation n^2-2n-28=0 are n=1+sqrt(29),n=1-sqrt(29)

Find the zeros of n^2-2n-28=0

The zeros of n^2-2n-28=0 are n=1+sqrt(29),n=1-sqrt(29)

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n^{2} - 2 n - 28 = 0

n = 1 + 29, n = 1 - 29

Solution steps

n^{2} - 2 n - 28 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 28) = 229

n_{1, 2} = \frac{- ( - 2 ) \pm 2 29}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 2 ) + 2 29}{2 \cdot 1}, n_{2} = \frac{- ( - 2 ) - 2 29}{2 \cdot 1}

n = \frac{- ( - 2 ) + 2 29}{2 \cdot 1} : 1 + 29

n = \frac{- ( - 2 ) - 2 29}{2 \cdot 1} : 1 - 29

The solutions to the quadratic equation are:

n = 1 + 29, n = 1 - 29

(x - 3) (x - 4) = 12

4 x^{2} - 20 x + 20 = 0

x^{2} + 3 x - 25 = - 7

3 x^{2} = x^{2} + 25

24 x - 12 x^{2} = 0

What are the solutions to the equation n^2-2n-28=0 ?
The solutions to the equation n^2-2n-28=0 are n=1+sqrt(29),n=1-sqrt(29)
Find the zeros of n^2-2n-28=0
The zeros of n^2-2n-28=0 are n=1+sqrt(29),n=1-sqrt(29)