What are the solutions to the equation 2x^2+x+7=0 ?

The solutions to the equation 2x^2+x+7=0 are x=-1/4+i(sqrt(55))/4 ,x=-1/4-i(sqrt(55))/4

Find the zeros of 2x^2+x+7=0

The zeros of 2x^2+x+7=0 are x=-1/4+i(sqrt(55))/4 ,x=-1/4-i(sqrt(55))/4

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2 x^{2} + x + 7 = 0

x = - \frac{1}{4} + i \frac{55}{4}, x = - \frac{1}{4} - i \frac{55}{4}

Solution steps

2 x^{2} + x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

Simplify

1^{2} - 4 \cdot 2 \cdot 7 : 55 i

x_{1, 2} = \frac{- 1 \pm 55 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 55 i}{2 \cdot 2}, x_{2} = \frac{- 1 - 55 i}{2 \cdot 2}

x = \frac{- 1 + 55 i}{2 \cdot 2} : - \frac{1}{4} + i \frac{55}{4}

x = \frac{- 1 - 55 i}{2 \cdot 2} : - \frac{1}{4} - i \frac{55}{4}

The solutions to the quadratic equation are:

x = - \frac{1}{4} + i \frac{55}{4}, x = - \frac{1}{4} - i \frac{55}{4}

\frac{1 + 43 i}{- 5 + 7 i}

∣ 2 x ∣ < 8

\frac{4}{6} = \frac{x}{9}

\frac{p}{4} = - 13

∣ 2 x - 3 ∣ \leq 7

What are the solutions to the equation 2x^2+x+7=0 ?
The solutions to the equation 2x^2+x+7=0 are x=-1/4+i(sqrt(55))/4 ,x=-1/4-i(sqrt(55))/4
Find the zeros of 2x^2+x+7=0
The zeros of 2x^2+x+7=0 are x=-1/4+i(sqrt(55))/4 ,x=-1/4-i(sqrt(55))/4