What are the solutions to the equation 8p^2+6p-48=0 ?

The solutions to the equation 8p^2+6p-48=0 are p=(-3+sqrt(393))/8 ,p=-(3+sqrt(393))/8

Find the zeros of 8p^2+6p-48=0

The zeros of 8p^2+6p-48=0 are p=(-3+sqrt(393))/8 ,p=-(3+sqrt(393))/8

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8 p^{2} + 6 p - 48 = 0

p = \frac{- 3 + 393}{8}, p = - \frac{3 + 393}{8}

Decimal Notation

p = 2.10302 \dots, p = - 2.85302 \dots

Solution steps

8 p^{2} + 6 p - 48 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 8 ( - 48 )}{2 \cdot 8}

6^{2} - 4 \cdot 8 (- 48) = 2393

p_{1, 2} = \frac{- 6 \pm 2 393}{2 \cdot 8}

Separate the solutions

p_{1} = \frac{- 6 + 2 393}{2 \cdot 8}, p_{2} = \frac{- 6 - 2 393}{2 \cdot 8}

p = \frac{- 6 + 2 393}{2 \cdot 8} : \frac{- 3 + 393}{8}

p = \frac{- 6 - 2 393}{2 \cdot 8} : - \frac{3 + 393}{8}

The solutions to the quadratic equation are:

p = \frac{- 3 + 393}{8}, p = - \frac{3 + 393}{8}

λ^{2} + 2 λ - 80 = 0

6 x^{2} - 5 x - 50 = 0

2 x^{2} - 6 x + 3 = 0

0.2 (2 x - 6) (8 x - 6) = 0

x, y + 5 x y = x^{2}

What are the solutions to the equation 8p^2+6p-48=0 ?
The solutions to the equation 8p^2+6p-48=0 are p=(-3+sqrt(393))/8 ,p=-(3+sqrt(393))/8
Find the zeros of 8p^2+6p-48=0
The zeros of 8p^2+6p-48=0 are p=(-3+sqrt(393))/8 ,p=-(3+sqrt(393))/8