What are the solutions to the equation 2x^2-4x-9=0 ?

The solutions to the equation 2x^2-4x-9=0 are x=(2+sqrt(22))/2 ,x=(2-sqrt(22))/2

Find the zeros of 2x^2-4x-9=0

The zeros of 2x^2-4x-9=0 are x=(2+sqrt(22))/2 ,x=(2-sqrt(22))/2

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2 x^{2} - 4 x - 9 = 0

x = \frac{2 + 22}{2}, x = \frac{2 - 22}{2}

Decimal Notation

x = 3.34520 \dots, x = - 1.34520 \dots

Solution steps

2 x^{2} - 4 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 (- 9) = 222

x_{1, 2} = \frac{- ( - 4 ) \pm 2 22}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 22}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 2 22}{2 \cdot 2}

x = \frac{- ( - 4 ) + 2 22}{2 \cdot 2} : \frac{2 + 22}{2}

x = \frac{- ( - 4 ) - 2 22}{2 \cdot 2} : \frac{2 - 22}{2}

The solutions to the quadratic equation are:

x = \frac{2 + 22}{2}, x = \frac{2 - 22}{2}

((\frac{1}{16})^{- \frac{2}{3}})^{\frac{3}{4}}

- 5 (1 - 5 x) + 5 (- 8 x - 2) = 4 x - 8 x

1 + \frac{x - 1}{2} = 3 x

(4 x^{3} yz)^{3}

What are the solutions to the equation 2x^2-4x-9=0 ?
The solutions to the equation 2x^2-4x-9=0 are x=(2+sqrt(22))/2 ,x=(2-sqrt(22))/2
Find the zeros of 2x^2-4x-9=0
The zeros of 2x^2-4x-9=0 are x=(2+sqrt(22))/2 ,x=(2-sqrt(22))/2