What are the solutions to the equation 9x^2+6x-18=0 ?

The solutions to the equation 9x^2+6x-18=0 are x=(-1+sqrt(19))/3 ,x=-(1+sqrt(19))/3

Find the zeros of 9x^2+6x-18=0

The zeros of 9x^2+6x-18=0 are x=(-1+sqrt(19))/3 ,x=-(1+sqrt(19))/3

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9 x^{2} + 6 x - 18 = 0

x = \frac{- 1 + 19}{3}, x = - \frac{1 + 19}{3}

Decimal Notation

x = 1.11963 \dots, x = - 1.78629 \dots

Solution steps

9 x^{2} + 6 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 9 ( - 18 )}{2 \cdot 9}

6^{2} - 4 \cdot 9 (- 18) = 619

x_{1, 2} = \frac{- 6 \pm 6 19}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 6 + 6 19}{2 \cdot 9}, x_{2} = \frac{- 6 - 6 19}{2 \cdot 9}

x = \frac{- 6 + 6 19}{2 \cdot 9} : \frac{- 1 + 19}{3}

x = \frac{- 6 - 6 19}{2 \cdot 9} : - \frac{1 + 19}{3}

The solutions to the quadratic equation are:

x = \frac{- 1 + 19}{3}, x = - \frac{1 + 19}{3}

x^{2} - 6 = 16 x + 30

x^{2} - 26 = 0

7 x^{2} - 84 x = 0

5 x^{2} = 120

x^{2} + (- \frac{3}{7})^{2} = 1

What are the solutions to the equation 9x^2+6x-18=0 ?
The solutions to the equation 9x^2+6x-18=0 are x=(-1+sqrt(19))/3 ,x=-(1+sqrt(19))/3
Find the zeros of 9x^2+6x-18=0
The zeros of 9x^2+6x-18=0 are x=(-1+sqrt(19))/3 ,x=-(1+sqrt(19))/3