What are the solutions to the equation 3x^2+8x+40=0 ?

The solutions to the equation 3x^2+8x+40=0 are x=-4/3+i(2sqrt(26))/3 ,x=-4/3-i(2sqrt(26))/3

Find the zeros of 3x^2+8x+40=0

The zeros of 3x^2+8x+40=0 are x=-4/3+i(2sqrt(26))/3 ,x=-4/3-i(2sqrt(26))/3

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3 x^{2} + 8 x + 40 = 0

x = - \frac{4}{3} + i \frac{2 26}{3}, x = - \frac{4}{3} - i \frac{2 26}{3}

Solution steps

3 x^{2} + 8 x + 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 3 \cdot 40}{2 \cdot 3}

Simplify

8^{2} - 4 \cdot 3 \cdot 40 : 426 i

x_{1, 2} = \frac{- 8 \pm 4 26 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 8 + 4 26 i}{2 \cdot 3}, x_{2} = \frac{- 8 - 4 26 i}{2 \cdot 3}

x = \frac{- 8 + 4 26 i}{2 \cdot 3} : - \frac{4}{3} + i \frac{2 26}{3}

x = \frac{- 8 - 4 26 i}{2 \cdot 3} : - \frac{4}{3} - i \frac{2 26}{3}

The solutions to the quadratic equation are:

x = - \frac{4}{3} + i \frac{2 26}{3}, x = - \frac{4}{3} - i \frac{2 26}{3}

x = y^{2} + 4 y, y

- x^{2} + 5 x - 2

- x^{2} + 5 x + 3

x, x^{2} y + y^{2} = k

2 x^{2} - 3 x - 16 = 16

What are the solutions to the equation 3x^2+8x+40=0 ?
The solutions to the equation 3x^2+8x+40=0 are x=-4/3+i(2sqrt(26))/3 ,x=-4/3-i(2sqrt(26))/3
Find the zeros of 3x^2+8x+40=0
The zeros of 3x^2+8x+40=0 are x=-4/3+i(2sqrt(26))/3 ,x=-4/3-i(2sqrt(26))/3