What are the solutions to the equation p^2+0.2p-0.96=0 ?

The solutions to the equation p^2+0.2p-0.96=0 are p=(-1+sqrt(97))/(10),p=-(1+sqrt(97))/(10)

Find the zeros of p^2+0.2p-0.96=0

The zeros of p^2+0.2p-0.96=0 are p=(-1+sqrt(97))/(10),p=-(1+sqrt(97))/(10)

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p^{2} + 0.2 p - 0.96 = 0

p = \frac{- 1 + 97}{10}, p = - \frac{1 + 97}{10}

Solution steps

p^{2} + 0.2 p - 0.96 = 0

Multiply both sides by

100

100 p^{2} + 20 p - 96 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 100 ( - 96 )}{2 \cdot 100}

2 0^{2} - 4 \cdot 100 (- 96) = 2097

p_{1, 2} = \frac{- 20 \pm 20 97}{2 \cdot 100}

Separate the solutions

p_{1} = \frac{- 20 + 20 97}{2 \cdot 100}, p_{2} = \frac{- 20 - 20 97}{2 \cdot 100}

p = \frac{- 20 + 20 97}{2 \cdot 100} : \frac{- 1 + 97}{10}

p = \frac{- 20 - 20 97}{2 \cdot 100} : - \frac{1 + 97}{10}

The solutions to the quadratic equation are:

p = \frac{- 1 + 97}{10}, p = - \frac{1 + 97}{10}

v^{2} = 256

(19 x - 2 x) (35 - 2 x) = 465

2 (3 x)^{2} = 162

44 = 1.5 (\frac{2}{3}) v + 0.06 v^{2}

m + 1 = - 4 m^{2}

What are the solutions to the equation p^2+0.2p-0.96=0 ?
The solutions to the equation p^2+0.2p-0.96=0 are p=(-1+sqrt(97))/(10),p=-(1+sqrt(97))/(10)
Find the zeros of p^2+0.2p-0.96=0
The zeros of p^2+0.2p-0.96=0 are p=(-1+sqrt(97))/(10),p=-(1+sqrt(97))/(10)