What are the solutions to the equation 15p^2+80p+80=0 ?

The solutions to the equation 15p^2+80p+80=0 are p=-4/3 ,p=-4

Find the zeros of 15p^2+80p+80=0

The zeros of 15p^2+80p+80=0 are p=-4/3 ,p=-4

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15 p^{2} + 80 p + 80 = 0

p = - \frac{4}{3}, p = - 4

Decimal Notation

p = - 1.33333 \dots, p = - 4

Solution steps

15 p^{2} + 80 p + 80 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 80 \pm 8 0 ^{2} - 4 \cdot 15 \cdot 80}{2 \cdot 15}

8 0^{2} - 4 \cdot 15 \cdot 80 = 40

p_{1, 2} = \frac{- 80 \pm 40}{2 \cdot 15}

Separate the solutions

p_{1} = \frac{- 80 + 40}{2 \cdot 15}, p_{2} = \frac{- 80 - 40}{2 \cdot 15}

p = \frac{- 80 + 40}{2 \cdot 15} : - \frac{4}{3}

p = \frac{- 80 - 40}{2 \cdot 15} : - 4

The solutions to the quadratic equation are:

p = - \frac{4}{3}, p = - 4

7 x^{2} - 5 = 0

x^{2} - 2 x + 1 = 80

3 a^{2} - a + 2 = 7 a

x, y = a x^{2} + b x + c

x^{2} - 16 x - 38 = 0

What are the solutions to the equation 15p^2+80p+80=0 ?
The solutions to the equation 15p^2+80p+80=0 are p=-4/3 ,p=-4
Find the zeros of 15p^2+80p+80=0
The zeros of 15p^2+80p+80=0 are p=-4/3 ,p=-4