What are the solutions to the equation roots 18x^2+476x+2400 ?

The solutions to the equation roots 18x^2+476x+2400 are x=(-119+sqrt(3361))/9 ,x=-(119+sqrt(3361))/9

Find the zeros of roots 18x^2+476x+2400

The zeros of roots 18x^2+476x+2400 are x=(-119+sqrt(3361))/9 ,x=-(119+sqrt(3361))/9

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roots

18 x^{2} + 476 x + 2400

x = \frac{- 119 + 3361}{9}, x = - \frac{119 + 3361}{9}

Solution steps

18 x^{2} + 476 x + 2400

The roots are the intercepts with the x-axis

(y = 0)

18 x^{2} + 476 x + 2400 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 476 \pm 47 6 ^{2} - 4 \cdot 18 \cdot 2400}{2 \cdot 18}

47 6^{2} - 4 \cdot 18 \cdot 2400 = 43361

x_{1, 2} = \frac{- 476 \pm 4 3361}{2 \cdot 18}

Separate the solutions

x_{1} = \frac{- 476 + 4 3361}{2 \cdot 18}, x_{2} = \frac{- 476 - 4 3361}{2 \cdot 18}

x = \frac{- 476 + 4 3361}{2 \cdot 18} : \frac{- 119 + 3361}{9}

x = \frac{- 476 - 4 3361}{2 \cdot 18} : - \frac{119 + 3361}{9}

The solutions to the quadratic equation are:

x = \frac{- 119 + 3361}{9}, x = - \frac{119 + 3361}{9}

3 k^{2} + 6 k = 105

x^{2} + 4 x - 5

3 (3 x + 6)^{2} - 46 = 5

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

x, 3 x^{2} + 6 x + 1 = 0

What are the solutions to the equation roots 18x^2+476x+2400 ?
The solutions to the equation roots 18x^2+476x+2400 are x=(-119+sqrt(3361))/9 ,x=-(119+sqrt(3361))/9
Find the zeros of roots 18x^2+476x+2400
The zeros of roots 18x^2+476x+2400 are x=(-119+sqrt(3361))/9 ,x=-(119+sqrt(3361))/9