What are the solutions to the equation roots 9x^2+6x+4 ?

The solutions to the equation roots 9x^2+6x+4 are x=-1/3+i(sqrt(3))/3 ,x=-1/3-i(sqrt(3))/3

Find the zeros of roots 9x^2+6x+4

The zeros of roots 9x^2+6x+4 are x=-1/3+i(sqrt(3))/3 ,x=-1/3-i(sqrt(3))/3

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roots

9 x^{2} + 6 x + 4

x = - \frac{1}{3} + i \frac{3}{3}, x = - \frac{1}{3} - i \frac{3}{3}

Solution steps

9 x^{2} + 6 x + 4

The roots are the intercepts with the x-axis

(y = 0)

9 x^{2} + 6 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

Simplify

6^{2} - 4 \cdot 9 \cdot 4 : 63 i

x_{1, 2} = \frac{- 6 \pm 6 3 i}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 6 + 6 3 i}{2 \cdot 9}, x_{2} = \frac{- 6 - 6 3 i}{2 \cdot 9}

x = \frac{- 6 + 6 3 i}{2 \cdot 9} : - \frac{1}{3} + i \frac{3}{3}

x = \frac{- 6 - 6 3 i}{2 \cdot 9} : - \frac{1}{3} - i \frac{3}{3}

The solutions to the quadratic equation are:

x = - \frac{1}{3} + i \frac{3}{3}, x = - \frac{1}{3} - i \frac{3}{3}

x, z = 64 - x^{2} - y^{2}

2 x + c = 3 x^{2} - c x - 5, c

2 X^{2} + 7 X + 6 = 0

4 x^{2} + 8 x - 4.5 = 0

3 x^{2} - 84 x = - 117

What are the solutions to the equation roots 9x^2+6x+4 ?
The solutions to the equation roots 9x^2+6x+4 are x=-1/3+i(sqrt(3))/3 ,x=-1/3-i(sqrt(3))/3
Find the zeros of roots 9x^2+6x+4
The zeros of roots 9x^2+6x+4 are x=-1/3+i(sqrt(3))/3 ,x=-1/3-i(sqrt(3))/3