What are the solutions to the equation -p^2-3p-13=-2p^2 ?

The solutions to the equation -p^2-3p-13=-2p^2 are p=(3+sqrt(61))/2 ,p=(3-sqrt(61))/2

Find the zeros of -p^2-3p-13=-2p^2

The zeros of -p^2-3p-13=-2p^2 are p=(3+sqrt(61))/2 ,p=(3-sqrt(61))/2

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- p^{2} - 3 p - 13 = - 2 p^{2}

p = \frac{3 + 61}{2}, p = \frac{3 - 61}{2}

Decimal Notation

p = 5.40512 \dots, p = - 2.40512 \dots

Solution steps

- p^{2} - 3 p - 13 = - 2 p^{2}

Move

2 p^{2}

to the left side

p^{2} - 3 p - 13 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 13 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 13) = 61

p_{1, 2} = \frac{- ( - 3 ) \pm 61}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- ( - 3 ) + 61}{2 \cdot 1}, p_{2} = \frac{- ( - 3 ) - 61}{2 \cdot 1}

p = \frac{- ( - 3 ) + 61}{2 \cdot 1} : \frac{3 + 61}{2}

p = \frac{- ( - 3 ) - 61}{2 \cdot 1} : \frac{3 - 61}{2}

The solutions to the quadratic equation are:

p = \frac{3 + 61}{2}, p = \frac{3 - 61}{2}

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3 x^{2} + 15 x = - 22

30 + 27 x = 400 - x^{2}

What are the solutions to the equation -p^2-3p-13=-2p^2 ?
The solutions to the equation -p^2-3p-13=-2p^2 are p=(3+sqrt(61))/2 ,p=(3-sqrt(61))/2
Find the zeros of -p^2-3p-13=-2p^2
The zeros of -p^2-3p-13=-2p^2 are p=(3+sqrt(61))/2 ,p=(3-sqrt(61))/2