What are the solutions to the equation 2x^2+4x-7=0 ?

The solutions to the equation 2x^2+4x-7=0 are x=(-2+3sqrt(2))/2 ,x=-(sqrt(2)+3)/(sqrt(2))

Find the zeros of 2x^2+4x-7=0

The zeros of 2x^2+4x-7=0 are x=(-2+3sqrt(2))/2 ,x=-(sqrt(2)+3)/(sqrt(2))

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2 x^{2} + 4 x - 7 = 0

x = \frac{- 2 + 3 2}{2}, x = - \frac{2 + 3}{2}

Decimal Notation

x = 1.12132 \dots, x = - 3.12132 \dots

Solution steps

2 x^{2} + 4 x - 7 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 2 (- 7) = 62

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

Separate the solutions

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

x = \frac{- 4 + 6 2}{2 \cdot 2} : \frac{- 2 + 3 2}{2}

x = \frac{- 4 - 6 2}{2 \cdot 2} : - \frac{2 + 3}{2}

The solutions to the quadratic equation are:

x = \frac{- 2 + 3 2}{2}, x = - \frac{2 + 3}{2}

36 + 12 m^{2} + m^{4}

6 (x - 5) = 60

y^{2} - 4 yz + 4 z^{2}

3 x^{2} - 5 x = 2

a^{2} + 8 a + 15

What are the solutions to the equation 2x^2+4x-7=0 ?
The solutions to the equation 2x^2+4x-7=0 are x=(-2+3sqrt(2))/2 ,x=-(sqrt(2)+3)/(sqrt(2))
Find the zeros of 2x^2+4x-7=0
The zeros of 2x^2+4x-7=0 are x=(-2+3sqrt(2))/2 ,x=-(sqrt(2)+3)/(sqrt(2))