What are the solutions to the equation 3t^2+16t+19=-2t ?

The solutions to the equation 3t^2+16t+19=-2t are t=(-9+2sqrt(6))/3 ,t=-(9+2sqrt(6))/3

Find the zeros of 3t^2+16t+19=-2t

The zeros of 3t^2+16t+19=-2t are t=(-9+2sqrt(6))/3 ,t=-(9+2sqrt(6))/3

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3 t^{2} + 16 t + 19 = - 2 t

t = \frac{- 9 + 2 6}{3}, t = - \frac{9 + 2 6}{3}

Decimal Notation

t = - 1.36700 \dots, t = - 4.63299 \dots

Solution steps

3 t^{2} + 16 t + 19 = - 2 t

Move

2 t

to the left side

3 t^{2} + 18 t + 19 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 3 \cdot 19}{2 \cdot 3}

1 8^{2} - 4 \cdot 3 \cdot 19 = 46

t_{1, 2} = \frac{- 18 \pm 4 6}{2 \cdot 3}

Separate the solutions

t_{1} = \frac{- 18 + 4 6}{2 \cdot 3}, t_{2} = \frac{- 18 - 4 6}{2 \cdot 3}

t = \frac{- 18 + 4 6}{2 \cdot 3} : \frac{- 9 + 2 6}{3}

t = \frac{- 18 - 4 6}{2 \cdot 3} : - \frac{9 + 2 6}{3}

The solutions to the quadratic equation are:

t = \frac{- 9 + 2 6}{3}, t = - \frac{9 + 2 6}{3}

4 m^{2} + 8 m - 4 = 1

38 a^{4} b^{3} c^{2} + 12 a^{9} b^{8} - 34 a^{5} b c^{6}

9 (\frac{2}{3})^{1}

\frac{3}{4} (x + 8) = 9

\frac{\frac{x}{y} - \frac{y}{x}}{\frac{1}{5 x ^{2}} - \frac{1}{5 y ^{2}}}

What are the solutions to the equation 3t^2+16t+19=-2t ?
The solutions to the equation 3t^2+16t+19=-2t are t=(-9+2sqrt(6))/3 ,t=-(9+2sqrt(6))/3
Find the zeros of 3t^2+16t+19=-2t
The zeros of 3t^2+16t+19=-2t are t=(-9+2sqrt(6))/3 ,t=-(9+2sqrt(6))/3