What are the solutions to the equation p^2+p-42=0 ?

The solutions to the equation p^2+p-42=0 are p=6,p=-7

Find the zeros of p^2+p-42=0

The zeros of p^2+p-42=0 are p=6,p=-7

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p^{2} + p - 42 = 0

p = 6, p = - 7

Solution steps

p^{2} + p - 42 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 42 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 42) = 13

p_{1, 2} = \frac{- 1 \pm 13}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 1 + 13}{2 \cdot 1}, p_{2} = \frac{- 1 - 13}{2 \cdot 1}

p = \frac{- 1 + 13}{2 \cdot 1} : 6

p = \frac{- 1 - 13}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

p = 6, p = - 7

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

9 x^{2} - 6 x + 2

m^{2} + 10 m - 17 = 0

2 x^{2} - 44 x + 224 = 0

x^{2} - 8 x - 1 = 0

What are the solutions to the equation p^2+p-42=0 ?
The solutions to the equation p^2+p-42=0 are p=6,p=-7
Find the zeros of p^2+p-42=0
The zeros of p^2+p-42=0 are p=6,p=-7