What are the solutions to the equation 2n^2-4n-14=0 ?

The solutions to the equation 2n^2-4n-14=0 are n=1+2sqrt(2),n=1-2sqrt(2)

Find the zeros of 2n^2-4n-14=0

The zeros of 2n^2-4n-14=0 are n=1+2sqrt(2),n=1-2sqrt(2)

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2 n^{2} - 4 n - 14 = 0

n = 1 + 22, n = 1 - 22

Decimal Notation

n = 3.82842 \dots, n = - 1.82842 \dots

Solution steps

2 n^{2} - 4 n - 14 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 14 )}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 (- 14) = 82

n_{1, 2} = \frac{- ( - 4 ) \pm 8 2}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- ( - 4 ) + 8 2}{2 \cdot 2}, n_{2} = \frac{- ( - 4 ) - 8 2}{2 \cdot 2}

n = \frac{- ( - 4 ) + 8 2}{2 \cdot 2} : 1 + 22

n = \frac{- ( - 4 ) - 8 2}{2 \cdot 2} : 1 - 22

The solutions to the quadratic equation are:

n = 1 + 22, n = 1 - 22

- 4 i^{2}

- 2 x + 1 = - x + 8

3 x - 15

4 x + 134 = 8 x + 158

- (8 n - 2) = 3 + 10 (1 - 3 n)

What are the solutions to the equation 2n^2-4n-14=0 ?
The solutions to the equation 2n^2-4n-14=0 are n=1+2sqrt(2),n=1-2sqrt(2)
Find the zeros of 2n^2-4n-14=0
The zeros of 2n^2-4n-14=0 are n=1+2sqrt(2),n=1-2sqrt(2)