What are the solutions to the equation -2x^2+x-5=0 ?

The solutions to the equation -2x^2+x-5=0 are x= 1/4-i(sqrt(39))/4 ,x= 1/4+i(sqrt(39))/4

Find the zeros of -2x^2+x-5=0

The zeros of -2x^2+x-5=0 are x= 1/4-i(sqrt(39))/4 ,x= 1/4+i(sqrt(39))/4

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- 2 x^{2} + x - 5 = 0

x = \frac{1}{4} - i \frac{39}{4}, x = \frac{1}{4} + i \frac{39}{4}

Solution steps

- 2 x^{2} + x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) ( - 5 )}{2 ( - 2 )}

Simplify

1^{2} - 4 (- 2) (- 5) : 39 i

x_{1, 2} = \frac{- 1 \pm 39 i}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 1 + 39 i}{2 ( - 2 )}, x_{2} = \frac{- 1 - 39 i}{2 ( - 2 )}

x = \frac{- 1 + 39 i}{2 ( - 2 )} : \frac{1}{4} - i \frac{39}{4}

x = \frac{- 1 - 39 i}{2 ( - 2 )} : \frac{1}{4} + i \frac{39}{4}

The solutions to the quadratic equation are:

x = \frac{1}{4} - i \frac{39}{4}, x = \frac{1}{4} + i \frac{39}{4}

(\frac{- 3 uv}{w ^{5}})^{2}

5 n^{2} + 24 n - 5

- 4 + 2

(2 x^{2} y^{3})^{2}

6 (p - \frac{1}{3}) - \frac{5 - 3 p}{24} = \frac{5 + 2 p}{6}

What are the solutions to the equation -2x^2+x-5=0 ?
The solutions to the equation -2x^2+x-5=0 are x= 1/4-i(sqrt(39))/4 ,x= 1/4+i(sqrt(39))/4
Find the zeros of -2x^2+x-5=0
The zeros of -2x^2+x-5=0 are x= 1/4-i(sqrt(39))/4 ,x= 1/4+i(sqrt(39))/4