What are the solutions to the equation 2x^2+8x-128=0 ?

The solutions to the equation 2x^2+8x-128=0 are x=2(sqrt(17)-1),x=-2(1+sqrt(17))

Find the zeros of 2x^2+8x-128=0

The zeros of 2x^2+8x-128=0 are x=2(sqrt(17)-1),x=-2(1+sqrt(17))

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2 x^{2} + 8 x - 128 = 0

x = 2 (17 - 1), x = - 2 (1 + 17)

Decimal Notation

x = 6.24621 \dots, x = - 10.24621 \dots

Solution steps

2 x^{2} + 8 x - 128 = 0

Solve with the quadratic formula

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

8^{2} - 4 \cdot 2 (- 128) = 817

x_{1, 2} = \frac{- 8 \pm 8 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 8 17}{2 \cdot 2}, x_{2} = \frac{- 8 - 8 17}{2 \cdot 2}

x = \frac{- 8 + 8 17}{2 \cdot 2} : 2 (17 - 1)

x = \frac{- 8 - 8 17}{2 \cdot 2} : - 2 (1 + 17)

The solutions to the quadratic equation are:

x = 2 (17 - 1), x = - 2 (1 + 17)

x^{2} - 119 = 10 (x - 8)

x^{2} + \frac{20}{3} x + 64 = 0

x (x + 2) = 224

y, y^{2} = (x + 1) (x - 1) (x - 2)

\frac{3 x ^{2} - 248 x}{4} + 961 = 0

What are the solutions to the equation 2x^2+8x-128=0 ?
The solutions to the equation 2x^2+8x-128=0 are x=2(sqrt(17)-1),x=-2(1+sqrt(17))
Find the zeros of 2x^2+8x-128=0
The zeros of 2x^2+8x-128=0 are x=2(sqrt(17)-1),x=-2(1+sqrt(17))