What are the solutions to the equation 2x^2+x+10=0 ?

The solutions to the equation 2x^2+x+10=0 are x=-1/4+i(sqrt(79))/4 ,x=-1/4-i(sqrt(79))/4

Find the zeros of 2x^2+x+10=0

The zeros of 2x^2+x+10=0 are x=-1/4+i(sqrt(79))/4 ,x=-1/4-i(sqrt(79))/4

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2 x^{2} + x + 10 = 0

x = - \frac{1}{4} + i \frac{79}{4}, x = - \frac{1}{4} - i \frac{79}{4}

Solution steps

2 x^{2} + x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 10}{2 \cdot 2}

Simplify

1^{2} - 4 \cdot 2 \cdot 10 : 79 i

x_{1, 2} = \frac{- 1 \pm 79 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 79 i}{2 \cdot 2}, x_{2} = \frac{- 1 - 79 i}{2 \cdot 2}

x = \frac{- 1 + 79 i}{2 \cdot 2} : - \frac{1}{4} + i \frac{79}{4}

x = \frac{- 1 - 79 i}{2 \cdot 2} : - \frac{1}{4} - i \frac{79}{4}

The solutions to the quadratic equation are:

x = - \frac{1}{4} + i \frac{79}{4}, x = - \frac{1}{4} - i \frac{79}{4}

x^{4} < x^{3} + 6 x^{2}

4 g^{2} - 1

312 5^{\frac{1}{5}}

z^{2} + 7 z - 18

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

What are the solutions to the equation 2x^2+x+10=0 ?
The solutions to the equation 2x^2+x+10=0 are x=-1/4+i(sqrt(79))/4 ,x=-1/4-i(sqrt(79))/4
Find the zeros of 2x^2+x+10=0
The zeros of 2x^2+x+10=0 are x=-1/4+i(sqrt(79))/4 ,x=-1/4-i(sqrt(79))/4