What are the solutions to the equation 2p^2+8p=-3 ?

The solutions to the equation 2p^2+8p=-3 are p=(-4+sqrt(10))/2 ,p=-(4+sqrt(10))/2

Find the zeros of 2p^2+8p=-3

The zeros of 2p^2+8p=-3 are p=(-4+sqrt(10))/2 ,p=-(4+sqrt(10))/2

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2 p^{2} + 8 p = - 3

p = \frac{- 4 + 10}{2}, p = - \frac{4 + 10}{2}

Decimal Notation

p = - 0.41886 \dots, p = - 3.58113 \dots

Solution steps

2 p^{2} + 8 p = - 3

Move

3

to the left side

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

Solve with the quadratic formula

p_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

8^{2} - 4 \cdot 2 \cdot 3 = 210

p_{1, 2} = \frac{- 8 \pm 2 10}{2 \cdot 2}

Separate the solutions

p_{1} = \frac{- 8 + 2 10}{2 \cdot 2}, p_{2} = \frac{- 8 - 2 10}{2 \cdot 2}

p = \frac{- 8 + 2 10}{2 \cdot 2} : \frac{- 4 + 10}{2}

p = \frac{- 8 - 2 10}{2 \cdot 2} : - \frac{4 + 10}{2}

The solutions to the quadratic equation are:

p = \frac{- 4 + 10}{2}, p = - \frac{4 + 10}{2}

8^{2} = 6^{2} + x^{2}

x (x - 7) + 12 = 0

2 (x + 5)^{2} - 32 = 0

2 x^{2} - (k - 1) x - k = 0

3 y^{2} = 81

What are the solutions to the equation 2p^2+8p=-3 ?
The solutions to the equation 2p^2+8p=-3 are p=(-4+sqrt(10))/2 ,p=-(4+sqrt(10))/2
Find the zeros of 2p^2+8p=-3
The zeros of 2p^2+8p=-3 are p=(-4+sqrt(10))/2 ,p=-(4+sqrt(10))/2