What are the solutions to the equation 4x^2-8x-4=0 ?

The solutions to the equation 4x^2-8x-4=0 are x=1+sqrt(2),x=1-sqrt(2)

Find the zeros of 4x^2-8x-4=0

The zeros of 4x^2-8x-4=0 are x=1+sqrt(2),x=1-sqrt(2)

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4 x^{2} - 8 x - 4 = 0

x = 1 + 2, x = 1 - 2

Decimal Notation

x = 2.41421 \dots, x = - 0.41421 \dots

Solution steps

4 x^{2} - 8 x - 4 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + 2, x = 1 - 2

x^{2} + 4 x + 85 = 0

x^{2} - 6 x = - x^{2} + 6 x

a, 9 a^{2} + 12 ab = 32 b^{2}

c^{2} = 15

x^{2} - \frac{2}{3} = 0

What are the solutions to the equation 4x^2-8x-4=0 ?
The solutions to the equation 4x^2-8x-4=0 are x=1+sqrt(2),x=1-sqrt(2)
Find the zeros of 4x^2-8x-4=0
The zeros of 4x^2-8x-4=0 are x=1+sqrt(2),x=1-sqrt(2)