What are the solutions to the equation 2x^2+8x+7=0 ?

The solutions to the equation 2x^2+8x+7=0 are x=(-4+sqrt(2))/2 ,x=-(4+sqrt(2))/2

Find the zeros of 2x^2+8x+7=0

The zeros of 2x^2+8x+7=0 are x=(-4+sqrt(2))/2 ,x=-(4+sqrt(2))/2

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2 x^{2} + 8 x + 7 = 0

x = \frac{- 4 + 2}{2}, x = - \frac{4 + 2}{2}

Decimal Notation

x = - 1.29289 \dots, x = - 2.70710 \dots

Solution steps

2 x^{2} + 8 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

8^{2} - 4 \cdot 2 \cdot 7 = 22

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{- 4 + 2}{2}, x = - \frac{4 + 2}{2}

x^{3} - x^{2} - 3 x + 3

- 3 x - 2 \leq 19

\frac{5 ( 97 )}{129} + \frac{5 ( 11 )}{30} + 1

2 \cdot (2 x - 7) - 4 \cdot (x + 1) = 3 \cdot (2 x + 5)

4 (\frac{1}{2})^{0}

What are the solutions to the equation 2x^2+8x+7=0 ?
The solutions to the equation 2x^2+8x+7=0 are x=(-4+sqrt(2))/2 ,x=-(4+sqrt(2))/2
Find the zeros of 2x^2+8x+7=0
The zeros of 2x^2+8x+7=0 are x=(-4+sqrt(2))/2 ,x=-(4+sqrt(2))/2