What are the solutions to the equation 3q^2+q-11=-4 ?

The solutions to the equation 3q^2+q-11=-4 are q=(-1+sqrt(85))/6 ,q=(-1-sqrt(85))/6

Find the zeros of 3q^2+q-11=-4

The zeros of 3q^2+q-11=-4 are q=(-1+sqrt(85))/6 ,q=(-1-sqrt(85))/6

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3 q^{2} + q - 11 = - 4

q = \frac{- 1 + 85}{6}, q = \frac{- 1 - 85}{6}

Decimal Notation

q = 1.36992 \dots, q = - 1.70325 \dots

Solution steps

3 q^{2} + q - 11 = - 4

Move

4

to the left side

3 q^{2} + q - 7 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 7 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 7) = 85

q_{1, 2} = \frac{- 1 \pm 85}{2 \cdot 3}

Separate the solutions

q_{1} = \frac{- 1 + 85}{2 \cdot 3}, q_{2} = \frac{- 1 - 85}{2 \cdot 3}

q = \frac{- 1 + 85}{2 \cdot 3} : \frac{- 1 + 85}{6}

q = \frac{- 1 - 85}{2 \cdot 3} : \frac{- 1 - 85}{6}

The solutions to the quadratic equation are:

q = \frac{- 1 + 85}{6}, q = \frac{- 1 - 85}{6}

(x + 24) \times (x + 2) = 420

0 = x^{2} + 5 x - 24

- 12 x - 9 = 4 x^{2}

x, z = 1 - x^{2} - y^{2}

- 8 x^{2} - 33 = - 9 x^{2} + 8 x

What are the solutions to the equation 3q^2+q-11=-4 ?
The solutions to the equation 3q^2+q-11=-4 are q=(-1+sqrt(85))/6 ,q=(-1-sqrt(85))/6
Find the zeros of 3q^2+q-11=-4
The zeros of 3q^2+q-11=-4 are q=(-1+sqrt(85))/6 ,q=(-1-sqrt(85))/6