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 = 3.56155 \dots, x = - 0.56155 \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}

9 x^{2} = 10 x + 2

3 x^{2} - 2 = 0

24 - 3 x = - 6 x + 22

1 x - 8 = - 3

4 x - 2 < 12

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