What are the solutions to the equation 2x^2-1x-2=0 ?

The solutions to the equation 2x^2-1x-2=0 are x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4

Find the zeros of 2x^2-1x-2=0

The zeros of 2x^2-1x-2=0 are x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4

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2 x^{2} - 1 x - 2 = 0

x = \frac{1 + 17}{4}, x = \frac{1 - 17}{4}

Decimal Notation

x = 1.28077 \dots, x = - 0.78077 \dots

Solution steps

2 x^{2} - 1 \cdot x - 2 = 0

Refine

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 2 (- 2) = 17

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

Separate the solutions

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

x = \frac{- ( - 1 ) + 17}{2 \cdot 2} : \frac{1 + 17}{4}

x = \frac{- ( - 1 ) - 17}{2 \cdot 2} : \frac{1 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 17}{4}, x = \frac{1 - 17}{4}

- \frac{6}{3}

24 3^{\frac{5}{2}}

- x^{2} + 6 x > 5

6 (x + 3) + 4 \geq 46 and 9 x - 4 < 8 x - 4

3 y^{3} - 9 x y^{4}

What are the solutions to the equation 2x^2-1x-2=0 ?
The solutions to the equation 2x^2-1x-2=0 are x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4
Find the zeros of 2x^2-1x-2=0
The zeros of 2x^2-1x-2=0 are x=(1+sqrt(17))/4 ,x=(1-sqrt(17))/4