What are the solutions to the equation 3z^2+17z+18=0 ?

The solutions to the equation 3z^2+17z+18=0 are z=(-17+sqrt(73))/6 ,z=(-17-sqrt(73))/6

Find the zeros of 3z^2+17z+18=0

The zeros of 3z^2+17z+18=0 are z=(-17+sqrt(73))/6 ,z=(-17-sqrt(73))/6

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3 z^{2} + 17 z + 18 = 0

z = \frac{- 17 + 73}{6}, z = \frac{- 17 - 73}{6}

Decimal Notation

z = - 1.40933 \dots, z = - 4.25733 \dots

Solution steps

3 z^{2} + 17 z + 18 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 \cdot 3 \cdot 18}{2 \cdot 3}

1 7^{2} - 4 \cdot 3 \cdot 18 = 73

z_{1, 2} = \frac{- 17 \pm 73}{2 \cdot 3}

Separate the solutions

z_{1} = \frac{- 17 + 73}{2 \cdot 3}, z_{2} = \frac{- 17 - 73}{2 \cdot 3}

z = \frac{- 17 + 73}{2 \cdot 3} : \frac{- 17 + 73}{6}

z = \frac{- 17 - 73}{2 \cdot 3} : \frac{- 17 - 73}{6}

The solutions to the quadratic equation are:

z = \frac{- 17 + 73}{6}, z = \frac{- 17 - 73}{6}

12 a^{2} b - 18 a b^{2} - 30 a b^{3}

- x^{2} + 1 > 4 x + 1

5 x^{2} + 30 x + 125 = 0

- 3 + 39 \geq 7 m + 5 m

- k + 78 = 98 - 20

What are the solutions to the equation 3z^2+17z+18=0 ?
The solutions to the equation 3z^2+17z+18=0 are z=(-17+sqrt(73))/6 ,z=(-17-sqrt(73))/6
Find the zeros of 3z^2+17z+18=0
The zeros of 3z^2+17z+18=0 are z=(-17+sqrt(73))/6 ,z=(-17-sqrt(73))/6