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

The solutions to the equation 2x^2+x-4=0 are x=(-1+sqrt(33))/4 ,x=(-1-sqrt(33))/4

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

The zeros of 2x^2+x-4=0 are x=(-1+sqrt(33))/4 ,x=(-1-sqrt(33))/4

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

x = \frac{- 1 + 33}{4}, x = \frac{- 1 - 33}{4}

Decimal Notation

x = 1.18614 \dots, x = - 1.68614 \dots

Solution steps

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 2 (- 4) = 33

x_{1, 2} = \frac{- 1 \pm 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 33}{2 \cdot 2}, x_{2} = \frac{- 1 - 33}{2 \cdot 2}

x = \frac{- 1 + 33}{2 \cdot 2} : \frac{- 1 + 33}{4}

x = \frac{- 1 - 33}{2 \cdot 2} : \frac{- 1 - 33}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 33}{4}, x = \frac{- 1 - 33}{4}

\frac{1}{2} - \frac{4}{5}

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

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

20 \cdot 5

\frac{x - 16}{6} \geq \frac{x - 8}{9} + \frac{1}{18}

What are the solutions to the equation 2x^2+x-4=0 ?
The solutions to the equation 2x^2+x-4=0 are x=(-1+sqrt(33))/4 ,x=(-1-sqrt(33))/4
Find the zeros of 2x^2+x-4=0
The zeros of 2x^2+x-4=0 are x=(-1+sqrt(33))/4 ,x=(-1-sqrt(33))/4