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 \cdot 2 \cdot 5}{2 \cdot 2}

Simplify

1^{2} - 4 \cdot 2 \cdot 5 : 39 i

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

Separate the solutions

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

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

x = \frac{- 1 - 39 i}{2 \cdot 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{4 2}{3}

4 - 7 x \leq 32

(- 2)^{4}

x - 4 \leq - 9

- 3 k - (- 8) + 2

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