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

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

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

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

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

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

Decimal Notation

q = - 0.11556 \dots, q = - 2.88443 \dots

Solution steps

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

Move

q

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

3 m^{2} + 20 m + 25 = 0

4 p^{2} + 6 p + 1 = 2 p

- 16 x^{2} + 14 x + 4 = 0

- x^{2} + 6 x = - 1

9^{2} + x^{2} = 1 8^{2}

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