What are the solutions to the equation 2x^2+6x-4=0 ?

The solutions to the equation 2x^2+6x-4=0 are x=(-3+sqrt(17))/2 ,x=-(3+sqrt(17))/2

Find the zeros of 2x^2+6x-4=0

The zeros of 2x^2+6x-4=0 are x=(-3+sqrt(17))/2 ,x=-(3+sqrt(17))/2

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2 x^{2} + 6 x - 4 = 0

x = \frac{- 3 + 17}{2}, x = - \frac{3 + 17}{2}

Decimal Notation

x = 0.56155 \dots, x = - 3.56155 \dots

Solution steps

2 x^{2} + 6 x - 4 = 0

Solve with the quadratic formula

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

6^{2} - 4 \cdot 2 (- 4) = 217

x_{1, 2} = \frac{- 6 \pm 2 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 6 + 2 17}{2 \cdot 2}, x_{2} = \frac{- 6 - 2 17}{2 \cdot 2}

x = \frac{- 6 + 2 17}{2 \cdot 2} : \frac{- 3 + 17}{2}

x = \frac{- 6 - 2 17}{2 \cdot 2} : - \frac{3 + 17}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 17}{2}, x = - \frac{3 + 17}{2}

x^{2} = 12 - x

\frac{14 x ^{8} y ^{7}}{7 x y ^{8}}

2 \cdot \frac{4}{3}

\frac{1}{16} + \frac{1}{3}

3 \leq x

What are the solutions to the equation 2x^2+6x-4=0 ?
The solutions to the equation 2x^2+6x-4=0 are x=(-3+sqrt(17))/2 ,x=-(3+sqrt(17))/2
Find the zeros of 2x^2+6x-4=0
The zeros of 2x^2+6x-4=0 are x=(-3+sqrt(17))/2 ,x=-(3+sqrt(17))/2