What are the solutions to the equation 2x^2+x+67=0 ?

The solutions to the equation 2x^2+x+67=0 are x=-1/4+i(sqrt(535))/4 ,x=-1/4-i(sqrt(535))/4

Find the zeros of 2x^2+x+67=0

The zeros of 2x^2+x+67=0 are x=-1/4+i(sqrt(535))/4 ,x=-1/4-i(sqrt(535))/4

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2 x^{2} + x + 67 = 0

x = - \frac{1}{4} + i \frac{535}{4}, x = - \frac{1}{4} - i \frac{535}{4}

Solution steps

2 x^{2} + x + 67 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 67}{2 \cdot 2}

Simplify

1^{2} - 4 \cdot 2 \cdot 67 : 535 i

x_{1, 2} = \frac{- 1 \pm 535 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 535 i}{2 \cdot 2}, x_{2} = \frac{- 1 - 535 i}{2 \cdot 2}

x = \frac{- 1 + 535 i}{2 \cdot 2} : - \frac{1}{4} + i \frac{535}{4}

x = \frac{- 1 - 535 i}{2 \cdot 2} : - \frac{1}{4} - i \frac{535}{4}

The solutions to the quadratic equation are:

x = - \frac{1}{4} + i \frac{535}{4}, x = - \frac{1}{4} - i \frac{535}{4}

2 (8 y - 6) \leq - 4 (3 - 4 y)

x - 5 < 10 < x + 2

3 - \frac{3}{5}

∣ - x ∣ = 2

0 = 4 t^{2} + 9 + (2 t - 16)^{2}

What are the solutions to the equation 2x^2+x+67=0 ?
The solutions to the equation 2x^2+x+67=0 are x=-1/4+i(sqrt(535))/4 ,x=-1/4-i(sqrt(535))/4
Find the zeros of 2x^2+x+67=0
The zeros of 2x^2+x+67=0 are x=-1/4+i(sqrt(535))/4 ,x=-1/4-i(sqrt(535))/4