What are the solutions to the equation 9x^2+18x+4=0 ?

The solutions to the equation 9x^2+18x+4=0 are x=(-3+sqrt(5))/3 ,x=-(3+sqrt(5))/3

Find the zeros of 9x^2+18x+4=0

The zeros of 9x^2+18x+4=0 are x=(-3+sqrt(5))/3 ,x=-(3+sqrt(5))/3

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9 x^{2} + 18 x + 4 = 0

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

Decimal Notation

x = - 0.25464 \dots, x = - 1.74535 \dots

Solution steps

9 x^{2} + 18 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

1 8^{2} - 4 \cdot 9 \cdot 4 = 65

x_{1, 2} = \frac{- 18 \pm 6 5}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 18 + 6 5}{2 \cdot 9}, x_{2} = \frac{- 18 - 6 5}{2 \cdot 9}

x = \frac{- 18 + 6 5}{2 \cdot 9} : \frac{- 3 + 5}{3}

x = \frac{- 18 - 6 5}{2 \cdot 9} : - \frac{3 + 5}{3}

The solutions to the quadratic equation are:

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

x^{2} + 49 x = 0

2^{3} - [4 (5 - 3)^{3}] = - 8 x

15 x^{2} + 4 x - 4 = 0

\frac{x}{6} + 5 = \frac{1}{3} - x

\frac{4}{3} - 3 x < - 15

What are the solutions to the equation 9x^2+18x+4=0 ?
The solutions to the equation 9x^2+18x+4=0 are x=(-3+sqrt(5))/3 ,x=-(3+sqrt(5))/3
Find the zeros of 9x^2+18x+4=0
The zeros of 9x^2+18x+4=0 are x=(-3+sqrt(5))/3 ,x=-(3+sqrt(5))/3