What are the solutions to the equation 3x^2+18x-7=0 ?

The solutions to the equation 3x^2+18x-7=0 are x=(-9+sqrt(102))/3 ,x=-(9+sqrt(102))/3

Find the zeros of 3x^2+18x-7=0

The zeros of 3x^2+18x-7=0 are x=(-9+sqrt(102))/3 ,x=-(9+sqrt(102))/3

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3 x^{2} + 18 x - 7 = 0

x = \frac{- 9 + 102}{3}, x = - \frac{9 + 102}{3}

Decimal Notation

x = 0.36650 \dots, x = - 6.36650 \dots

Solution steps

3 x^{2} + 18 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 3 ( - 7 )}{2 \cdot 3}

1 8^{2} - 4 \cdot 3 (- 7) = 2102

x_{1, 2} = \frac{- 18 \pm 2 102}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 18 + 2 102}{2 \cdot 3}, x_{2} = \frac{- 18 - 2 102}{2 \cdot 3}

x = \frac{- 18 + 2 102}{2 \cdot 3} : \frac{- 9 + 102}{3}

x = \frac{- 18 - 2 102}{2 \cdot 3} : - \frac{9 + 102}{3}

The solutions to the quadratic equation are:

x = \frac{- 9 + 102}{3}, x = - \frac{9 + 102}{3}

(3 x + 2)^{2} = 121

9 (2 x - 1)^{2} - 8 = 0

2 (x^{2} + 4) = 10

\frac{x ^{2} - 8 x - 2}{3} = \frac{x ^{2} - 3 x + 2}{2}

x^{2} + 4 x + 13

What are the solutions to the equation 3x^2+18x-7=0 ?
The solutions to the equation 3x^2+18x-7=0 are x=(-9+sqrt(102))/3 ,x=-(9+sqrt(102))/3
Find the zeros of 3x^2+18x-7=0
The zeros of 3x^2+18x-7=0 are x=(-9+sqrt(102))/3 ,x=-(9+sqrt(102))/3