What are the solutions to the equation 12p^2+32p-4=8 ?

The solutions to the equation 12p^2+32p-4=8 are p= 1/3 ,p=-3

Find the zeros of 12p^2+32p-4=8

The zeros of 12p^2+32p-4=8 are p= 1/3 ,p=-3

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12 p^{2} + 32 p - 4 = 8

p = \frac{1}{3}, p = - 3

Decimal Notation

p = 0.33333 \dots, p = - 3

Solution steps

12 p^{2} + 32 p - 4 = 8

Move

8

to the left side

12 p^{2} + 32 p - 12 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 \cdot 12 ( - 12 )}{2 \cdot 12}

3 2^{2} - 4 \cdot 12 (- 12) = 40

p_{1, 2} = \frac{- 32 \pm 40}{2 \cdot 12}

Separate the solutions

p_{1} = \frac{- 32 + 40}{2 \cdot 12}, p_{2} = \frac{- 32 - 40}{2 \cdot 12}

p = \frac{- 32 + 40}{2 \cdot 12} : \frac{1}{3}

p = \frac{- 32 - 40}{2 \cdot 12} : - 3

The solutions to the quadratic equation are:

p = \frac{1}{3}, p = - 3

5 x^{2} - 30 x = 5

(x + 2) (x^{2} - 12 x + 35)

100 x - x^{3} \geq 0

- ∣ 1 + x ∣ + ∣ 1 - x ∣ = 2

9 - x \geq 10

What are the solutions to the equation 12p^2+32p-4=8 ?
The solutions to the equation 12p^2+32p-4=8 are p= 1/3 ,p=-3
Find the zeros of 12p^2+32p-4=8
The zeros of 12p^2+32p-4=8 are p= 1/3 ,p=-3