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 ( - 2 ) \cdot 4}{2 ( - 2 )}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

- 5 (4 x - 7) + 8 = 3 x

(2^{3})^{- 2}

3 (2 x + 5) + 33 \leq 0

\frac{2 x ^{4} y ^{2} \cdot y x ^{- 3}}{( 2 x ^{2} y ^{2} \cdot y x ^{- 1} ) ^{- 3}}

x^{2} + 4 x + 29 = 0

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