What are the solutions to the equation 2x^2+128x-4096=0 ?

The solutions to the equation 2x^2+128x-4096=0 are x=32(sqrt(3)-1),x=-32(1+sqrt(3))

Find the zeros of 2x^2+128x-4096=0

The zeros of 2x^2+128x-4096=0 are x=32(sqrt(3)-1),x=-32(1+sqrt(3))

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2 x^{2} + 128 x - 4096 = 0

x = 32 (3 - 1), x = - 32 (1 + 3)

Solution steps

2 x^{2} + 128 x - 4096 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 \cdot 2 ( - 4096 )}{2 \cdot 2}

12 8^{2} - 4 \cdot 2 (- 4096) = 1283

x_{1, 2} = \frac{- 128 \pm 128 3}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 128 + 128 3}{2 \cdot 2}, x_{2} = \frac{- 128 - 128 3}{2 \cdot 2}

x = \frac{- 128 + 128 3}{2 \cdot 2} : 32 (3 - 1)

x = \frac{- 128 - 128 3}{2 \cdot 2} : - 32 (1 + 3)

The solutions to the quadratic equation are:

x = 32 (3 - 1), x = - 32 (1 + 3)

4 (7 - 7 x^{2}) = 7

2 x^{2} + x - 3 = 0

x (x - 7) = 2 x

x (x - c) = a

9.81 t^{2} - 24 t - 140 = 0

What are the solutions to the equation 2x^2+128x-4096=0 ?
The solutions to the equation 2x^2+128x-4096=0 are x=32(sqrt(3)-1),x=-32(1+sqrt(3))
Find the zeros of 2x^2+128x-4096=0
The zeros of 2x^2+128x-4096=0 are x=32(sqrt(3)-1),x=-32(1+sqrt(3))