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(2),x=-2-sqrt(2)

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

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

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

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

Decimal Notation

x = - 0.58578 \dots, x = - 3.41421 \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 \cdot 4}{2 \cdot 2}

8^{2} - 4 \cdot 2 \cdot 4 = 42

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

∣ 2 x - 4 ∣ \leq 12

y^{2} - 2 y - 24 = 0

4 x - 3 + 2 x = 5 (2 x - 3)

\frac{1}{2} + \frac{1}{4} - \frac{2}{6} - \frac{3}{8}

c = 180 (6) + 250

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