What are the solutions to the equation 3x^2+4x+7=-1 ?

The solutions to the equation 3x^2+4x+7=-1 are x=-2/3+i(2sqrt(5))/3 ,x=-2/3-i(2sqrt(5))/3

Find the zeros of 3x^2+4x+7=-1

The zeros of 3x^2+4x+7=-1 are x=-2/3+i(2sqrt(5))/3 ,x=-2/3-i(2sqrt(5))/3

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3 x^{2} + 4 x + 7 = - 1

x = - \frac{2}{3} + i \frac{2 5}{3}, x = - \frac{2}{3} - i \frac{2 5}{3}

Solution steps

3 x^{2} + 4 x + 7 = - 1

Move

1

to the left side

3 x^{2} + 4 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 3 \cdot 8}{2 \cdot 3}

Simplify

4^{2} - 4 \cdot 3 \cdot 8 : 45 i

x_{1, 2} = \frac{- 4 \pm 4 5 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 4 + 4 5 i}{2 \cdot 3}, x_{2} = \frac{- 4 - 4 5 i}{2 \cdot 3}

x = \frac{- 4 + 4 5 i}{2 \cdot 3} : - \frac{2}{3} + i \frac{2 5}{3}

x = \frac{- 4 - 4 5 i}{2 \cdot 3} : - \frac{2}{3} - i \frac{2 5}{3}

The solutions to the quadratic equation are:

x = - \frac{2}{3} + i \frac{2 5}{3}, x = - \frac{2}{3} - i \frac{2 5}{3}

x^{2} - \frac{1}{2} x + 1

(x + 1)^{2} = 11

x^{2} + 13 = 17

- x^{2} + 12 x - 36 = 0

x^{2} + 4 x + 7 = 15

What are the solutions to the equation 3x^2+4x+7=-1 ?
The solutions to the equation 3x^2+4x+7=-1 are x=-2/3+i(2sqrt(5))/3 ,x=-2/3-i(2sqrt(5))/3
Find the zeros of 3x^2+4x+7=-1
The zeros of 3x^2+4x+7=-1 are x=-2/3+i(2sqrt(5))/3 ,x=-2/3-i(2sqrt(5))/3