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

The solutions to the equation 2x^2+8x-4=0 are x=-2+sqrt(6),x=-2-sqrt(6)

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

The zeros of 2x^2+8x-4=0 are x=-2+sqrt(6),x=-2-sqrt(6)

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

x = - 2 + 6, x = - 2 - 6

Decimal Notation

x = 0.44948 \dots, x = - 4.44948 \dots

Solution steps

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

Solve with the quadratic formula

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

8^{2} - 4 \cdot 2 (- 4) = 46

x_{1, 2} = \frac{- 8 \pm 4 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 4 6}{2 \cdot 2}, x_{2} = \frac{- 8 - 4 6}{2 \cdot 2}

x = \frac{- 8 + 4 6}{2 \cdot 2} : - 2 + 6

x = \frac{- 8 - 4 6}{2 \cdot 2} : - 2 - 6

The solutions to the quadratic equation are:

x = - 2 + 6, x = - 2 - 6

\frac{3}{4 ( \frac{1}{4} + 2 ) ^{\frac{5}{2}}}

10 x + 5 < - 8 + 6 x

\frac{1}{2 e ^{- 3}}

x 2 + x + 72

2 u - 9 < 8 u - 21

What are the solutions to the equation 2x^2+8x-4=0 ?
The solutions to the equation 2x^2+8x-4=0 are x=-2+sqrt(6),x=-2-sqrt(6)
Find the zeros of 2x^2+8x-4=0
The zeros of 2x^2+8x-4=0 are x=-2+sqrt(6),x=-2-sqrt(6)