What are the solutions to the equation 9z^2+6z+1=0 ?

The solutions to the equation 9z^2+6z+1=0 are z=-1/3

Find the zeros of 9z^2+6z+1=0

The zeros of 9z^2+6z+1=0 are z=-1/3

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9 z^{2} + 6 z + 1 = 0

z = - \frac{1}{3}

Decimal Notation

z = - 0.33333 \dots

Solution steps

9 z^{2} + 6 z + 1 = 0

Solve with the quadratic formula

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

6^{2} - 4 \cdot 9 \cdot 1 = 0

z_{1, 2} = \frac{- 6 \pm 0}{2 \cdot 9}

z = \frac{- 6}{2 \cdot 9}

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

z = - \frac{1}{3}

The solution to the quadratic equation is:

z = - \frac{1}{3}

x^{3} + x^{2} + 5 x + 5

x^{4} > x^{2}

\frac{2}{5} + \frac{3}{4}

(m + 4) (m - 5) = - 14

(x - 6)^{2} = 2

What are the solutions to the equation 9z^2+6z+1=0 ?
The solutions to the equation 9z^2+6z+1=0 are z=-1/3
Find the zeros of 9z^2+6z+1=0
The zeros of 9z^2+6z+1=0 are z=-1/3