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

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

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

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

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

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

Decimal Notation

x = 4.82842 \dots, x = - 0.82842 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 2 (- 8) = 82

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} - 8 x + 12 = 0

8.65 = \frac{x ^{2} \cdot 4 π \cdot 4}{1.18 \cdot 344 \cdot 2}

2 x^{2} - (a + 6) x + 3 a = 0

9 x^{2} - 7 x - 9 = 0

x^{2} - x + 7

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