What are the solutions to the equation p^2+9p+3=-4p^2 ?

The solutions to the equation p^2+9p+3=-4p^2 are p=(-9+sqrt(21))/(10),p=(-9-sqrt(21))/(10)

Find the zeros of p^2+9p+3=-4p^2

The zeros of p^2+9p+3=-4p^2 are p=(-9+sqrt(21))/(10),p=(-9-sqrt(21))/(10)

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p^{2} + 9 p + 3 = - 4 p^{2}

p = \frac{- 9 + 21}{10}, p = \frac{- 9 - 21}{10}

Decimal Notation

p = - 0.44174 \dots, p = - 1.35825 \dots

Solution steps

p^{2} + 9 p + 3 = - 4 p^{2}

Move

4 p^{2}

to the left side

5 p^{2} + 9 p + 3 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 5 \cdot 3}{2 \cdot 5}

9^{2} - 4 \cdot 5 \cdot 3 = 21

p_{1, 2} = \frac{- 9 \pm 21}{2 \cdot 5}

Separate the solutions

p_{1} = \frac{- 9 + 21}{2 \cdot 5}, p_{2} = \frac{- 9 - 21}{2 \cdot 5}

p = \frac{- 9 + 21}{2 \cdot 5} : \frac{- 9 + 21}{10}

p = \frac{- 9 - 21}{2 \cdot 5} : \frac{- 9 - 21}{10}

The solutions to the quadratic equation are:

p = \frac{- 9 + 21}{10}, p = \frac{- 9 - 21}{10}

y^{2} + 8.19 y - 15 = 0

3 (x + 11)^{2} + 12 = 0

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

1 x + 3 x + 1 = 8 x - (- -) 3

(x + 2) \cdot (x - 4) = x + 2

What are the solutions to the equation p^2+9p+3=-4p^2 ?
The solutions to the equation p^2+9p+3=-4p^2 are p=(-9+sqrt(21))/(10),p=(-9-sqrt(21))/(10)
Find the zeros of p^2+9p+3=-4p^2
The zeros of p^2+9p+3=-4p^2 are p=(-9+sqrt(21))/(10),p=(-9-sqrt(21))/(10)