What are the solutions to the equation 7x^2+2x+3=0 ?

The solutions to the equation 7x^2+2x+3=0 are x=-1/7+i(2sqrt(5))/7 ,x=-1/7-i(2sqrt(5))/7

Find the zeros of 7x^2+2x+3=0

The zeros of 7x^2+2x+3=0 are x=-1/7+i(2sqrt(5))/7 ,x=-1/7-i(2sqrt(5))/7

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7 x^{2} + 2 x + 3 = 0

x = - \frac{1}{7} + i \frac{2 5}{7}, x = - \frac{1}{7} - i \frac{2 5}{7}

Solution steps

7 x^{2} + 2 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 7 \cdot 3}{2 \cdot 7}

Simplify

2^{2} - 4 \cdot 7 \cdot 3 : 45 i

x_{1, 2} = \frac{- 2 \pm 4 5 i}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- 2 + 4 5 i}{2 \cdot 7}, x_{2} = \frac{- 2 - 4 5 i}{2 \cdot 7}

x = \frac{- 2 + 4 5 i}{2 \cdot 7} : - \frac{1}{7} + i \frac{2 5}{7}

x = \frac{- 2 - 4 5 i}{2 \cdot 7} : - \frac{1}{7} - i \frac{2 5}{7}

The solutions to the quadratic equation are:

x = - \frac{1}{7} + i \frac{2 5}{7}, x = - \frac{1}{7} - i \frac{2 5}{7}

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- \frac{1}{2} (2 n + 4) + 6 = - 9 + 4 (2 n + 1)

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What are the solutions to the equation 7x^2+2x+3=0 ?
The solutions to the equation 7x^2+2x+3=0 are x=-1/7+i(2sqrt(5))/7 ,x=-1/7-i(2sqrt(5))/7
Find the zeros of 7x^2+2x+3=0
The zeros of 7x^2+2x+3=0 are x=-1/7+i(2sqrt(5))/7 ,x=-1/7-i(2sqrt(5))/7