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

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

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

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

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

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

Decimal Notation

n = - 0.31897 \dots, n = - 1.88102 \dots

Solution steps

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

Move

4 n^{2}

to the left side

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

Solve with the quadratic formula

n_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

20 - x^{2} = 10

x^{2} - 7 x + \frac{1}{4} = 0

x^{2} + 25 = - 24

3 (x - 2)^{2} = 27

5 x^{2} = - x + 7

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