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

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

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

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

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

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

Decimal Notation

m = - 0.31897 \dots, m = - 1.88102 \dots

Solution steps

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

Move

4 m^{2}

to the left side

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

Solve with the quadratic formula

m_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

(3 v - 2)^{2} = 5

(x + 2)^{2} = 2 x - 1

a^{1} = 5 x (3 x + 2)

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

2 x - x (2 - x) = 12

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