What are the solutions to the equation n^2+4n-63=5 ?

The solutions to the equation n^2+4n-63=5 are n=2(3sqrt(2)-1),n=-2(1+3sqrt(2))

Find the zeros of n^2+4n-63=5

The zeros of n^2+4n-63=5 are n=2(3sqrt(2)-1),n=-2(1+3sqrt(2))

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n^{2} + 4 n - 63 = 5

n = 2 (32 - 1), n = - 2 (1 + 32)

Decimal Notation

n = 6.48528 \dots, n = - 10.48528 \dots

Solution steps

n^{2} + 4 n - 63 = 5

Move

5

to the left side

n^{2} + 4 n - 68 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 68 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 68) = 122

n_{1, 2} = \frac{- 4 \pm 12 2}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 4 + 12 2}{2 \cdot 1}, n_{2} = \frac{- 4 - 12 2}{2 \cdot 1}

n = \frac{- 4 + 12 2}{2 \cdot 1} : 2 (32 - 1)

n = \frac{- 4 - 12 2}{2 \cdot 1} : - 2 (1 + 32)

The solutions to the quadratic equation are:

n = 2 (32 - 1), n = - 2 (1 + 32)

- 2 (x - 3) (x - 11) = 0

9 x^{2} + 33 x = - 30

x^{2} - 10 x = 26

1 + 2 + n = \frac{n ( n + 1 )}{2}

5 x^{2} + 8 x - 15 = 0

What are the solutions to the equation n^2+4n-63=5 ?
The solutions to the equation n^2+4n-63=5 are n=2(3sqrt(2)-1),n=-2(1+3sqrt(2))
Find the zeros of n^2+4n-63=5
The zeros of n^2+4n-63=5 are n=2(3sqrt(2)-1),n=-2(1+3sqrt(2))