What are the solutions to the equation 3x^2+19x+6=0 ?

The solutions to the equation 3x^2+19x+6=0 are x=-1/3 ,x=-6

Find the zeros of 3x^2+19x+6=0

The zeros of 3x^2+19x+6=0 are x=-1/3 ,x=-6

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3 x^{2} + 19 x + 6 = 0

x = - \frac{1}{3}, x = - 6

Decimal Notation

x = - 0.33333 \dots, x = - 6

Solution steps

3 x^{2} + 19 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

1 9^{2} - 4 \cdot 3 \cdot 6 = 17

x_{1, 2} = \frac{- 19 \pm 17}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 19 + 17}{2 \cdot 3}, x_{2} = \frac{- 19 - 17}{2 \cdot 3}

x = \frac{- 19 + 17}{2 \cdot 3} : - \frac{1}{3}

x = \frac{- 19 - 17}{2 \cdot 3} : - 6

The solutions to the quadratic equation are:

x = - \frac{1}{3}, x = - 6

x - (x - x) = x

- 6 + 3 x = 9

2 x \frac{3}{2} \leq - \frac{1}{2}

9 (2 x + 1) < 9 x - 18

\frac{4}{5} y^{2} - 64

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