What are the solutions to the equation 2n^2-8n=256 ?

The solutions to the equation 2n^2-8n=256 are n=2(1+sqrt(33)),n=2(1-sqrt(33))

Find the zeros of 2n^2-8n=256

The zeros of 2n^2-8n=256 are n=2(1+sqrt(33)),n=2(1-sqrt(33))

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2 n^{2} - 8 n = 256

n = 2 (1 + 33), n = 2 (1 - 33)

Decimal Notation

n = 13.48912 \dots, n = - 9.48912 \dots

Solution steps

2 n^{2} - 8 n = 256

Move

256

to the left side

2 n^{2} - 8 n - 256 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 ( - 256 )}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 (- 256) = 833

n_{1, 2} = \frac{- ( - 8 ) \pm 8 33}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- ( - 8 ) + 8 33}{2 \cdot 2}, n_{2} = \frac{- ( - 8 ) - 8 33}{2 \cdot 2}

n = \frac{- ( - 8 ) + 8 33}{2 \cdot 2} : 2 (1 + 33)

n = \frac{- ( - 8 ) - 8 33}{2 \cdot 2} : 2 (1 - 33)

The solutions to the quadratic equation are:

n = 2 (1 + 33), n = 2 (1 - 33)

4 x^{2} - 9 x + 6 = 0

2 x^{2} = - x

9 x^{2} - 15 x + 4 = 0

1 = x^{2} - 6 x

36 x^{2} = 0

What are the solutions to the equation 2n^2-8n=256 ?
The solutions to the equation 2n^2-8n=256 are n=2(1+sqrt(33)),n=2(1-sqrt(33))
Find the zeros of 2n^2-8n=256
The zeros of 2n^2-8n=256 are n=2(1+sqrt(33)),n=2(1-sqrt(33))