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

The solutions to the equation q^2+11q+3=-4q^2 are q=(-11+sqrt(61))/(10),q=(-11-sqrt(61))/(10)

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

The zeros of q^2+11q+3=-4q^2 are q=(-11+sqrt(61))/(10),q=(-11-sqrt(61))/(10)

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

q = \frac{- 11 + 61}{10}, q = \frac{- 11 - 61}{10}

Decimal Notation

q = - 0.31897 \dots, q = - 1.88102 \dots

Solution steps

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

Move

4 q^{2}

to the left side

5 q^{2} + 11 q + 3 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 5 \cdot 3}{2 \cdot 5}

1 1^{2} - 4 \cdot 5 \cdot 3 = 61

q_{1, 2} = \frac{- 11 \pm 61}{2 \cdot 5}

Separate the solutions

q_{1} = \frac{- 11 + 61}{2 \cdot 5}, q_{2} = \frac{- 11 - 61}{2 \cdot 5}

q = \frac{- 11 + 61}{2 \cdot 5} : \frac{- 11 + 61}{10}

q = \frac{- 11 - 61}{2 \cdot 5} : \frac{- 11 - 61}{10}

The solutions to the quadratic equation are:

q = \frac{- 11 + 61}{10}, q = \frac{- 11 - 61}{10}

2 x^{2} + 8 x + 3 = 0

3 x^{2} + 10 x + 9 = - 3 x + 5

6 0^{2} + 1 1^{2} = x^{2}

x, x y^{2} + x^{2} y = 2

6 x^{2} - 8 = 52

What are the solutions to the equation q^2+11q+3=-4q^2 ?
The solutions to the equation q^2+11q+3=-4q^2 are q=(-11+sqrt(61))/(10),q=(-11-sqrt(61))/(10)
Find the zeros of q^2+11q+3=-4q^2
The zeros of q^2+11q+3=-4q^2 are q=(-11+sqrt(61))/(10),q=(-11-sqrt(61))/(10)