What are the solutions to the equation 3x^2+19x+16=0 ?

The solutions to the equation 3x^2+19x+16=0 are x=-1,x=-16/3

Find the zeros of 3x^2+19x+16=0

The zeros of 3x^2+19x+16=0 are x=-1,x=-16/3

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3 x^{2} + 19 x + 16 = 0

x = - 1, x = - \frac{16}{3}

Decimal Notation

x = - 1, x = - 5.33333 \dots

Solution steps

3 x^{2} + 19 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 3 \cdot 16}{2 \cdot 3}

1 9^{2} - 4 \cdot 3 \cdot 16 = 13

x_{1, 2} = \frac{- 19 \pm 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 19 + 13}{2 \cdot 3}, x_{2} = \frac{- 19 - 13}{2 \cdot 3}

x = \frac{- 19 + 13}{2 \cdot 3} : - 1

x = \frac{- 19 - 13}{2 \cdot 3} : - \frac{16}{3}

The solutions to the quadratic equation are:

x = - 1, x = - \frac{16}{3}

15

\frac{x - 6}{x ^{2} + 3} \leq 0

x^{2} + x - 42 \leq 0

\frac{4}{20}

What are the solutions to the equation 3x^2+19x+16=0 ?
The solutions to the equation 3x^2+19x+16=0 are x=-1,x=-16/3
Find the zeros of 3x^2+19x+16=0
The zeros of 3x^2+19x+16=0 are x=-1,x=-16/3