What are the solutions to the equation 3t^2+10t+1=t ?

The solutions to the equation 3t^2+10t+1=t are t=(-9+sqrt(69))/6 ,t=(-9-sqrt(69))/6

Find the zeros of 3t^2+10t+1=t

The zeros of 3t^2+10t+1=t are t=(-9+sqrt(69))/6 ,t=(-9-sqrt(69))/6

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3 t^{2} + 10 t + 1 = t

t = \frac{- 9 + 69}{6}, t = \frac{- 9 - 69}{6}

Decimal Notation

t = - 0.11556 \dots, t = - 2.88443 \dots

Solution steps

3 t^{2} + 10 t + 1 = t

Move

t

to the left side

3 t^{2} + 9 t + 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

9^{2} - 4 \cdot 3 \cdot 1 = 69

t_{1, 2} = \frac{- 9 \pm 69}{2 \cdot 3}

Separate the solutions

t_{1} = \frac{- 9 + 69}{2 \cdot 3}, t_{2} = \frac{- 9 - 69}{2 \cdot 3}

t = \frac{- 9 + 69}{2 \cdot 3} : \frac{- 9 + 69}{6}

t = \frac{- 9 - 69}{2 \cdot 3} : \frac{- 9 - 69}{6}

The solutions to the quadratic equation are:

t = \frac{- 9 + 69}{6}, t = \frac{- 9 - 69}{6}

11 + 12 x^{2} = 10 - 15 x

16.1 x^{2} - 200 x - 200 = 0

x^{2} + 10 x = 2184

(x - 4) (2 - x) + 2 = 1 - x (3 - x)

p^{2} + 13 p + 19 = 0

What are the solutions to the equation 3t^2+10t+1=t ?
The solutions to the equation 3t^2+10t+1=t are t=(-9+sqrt(69))/6 ,t=(-9-sqrt(69))/6
Find the zeros of 3t^2+10t+1=t
The zeros of 3t^2+10t+1=t are t=(-9+sqrt(69))/6 ,t=(-9-sqrt(69))/6