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=(2-3sqrt(2))/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=(2-3sqrt(2))/2

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

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

Decimal Notation

x = 3.12132 \dots, x = - 1.12132 \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}{2}

The solutions to the quadratic equation are:

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

\frac{- 8}{9} + \frac{5}{7} - (\frac{4}{5} - \frac{- 2}{7} + \frac{13}{7})

- 7 < 8 + 3 x \leq 11

((\frac{2}{3})^{2})^{3}^{- 4}

(\frac{11}{12})^{2}

x^{2} + 10 x = 6

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=(2-3sqrt(2))/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=(2-3sqrt(2))/2