What are the solutions to the equation p^2+11p-12=0 ?

The solutions to the equation p^2+11p-12=0 are p=1,p=-12

Find the zeros of p^2+11p-12=0

The zeros of p^2+11p-12=0 are p=1,p=-12

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p^{2} + 11 p - 12 = 0

p = 1, p = - 12

Solution steps

p^{2} + 11 p - 12 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

1 1^{2} - 4 \cdot 1 \cdot (- 12) = 13

p_{1, 2} = \frac{- 11 \pm 13}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 11 + 13}{2 \cdot 1}, p_{2} = \frac{- 11 - 13}{2 \cdot 1}

p = \frac{- 11 + 13}{2 \cdot 1} : 1

p = \frac{- 11 - 13}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

p = 1, p = - 12

- 5 \leq \frac{x - 2}{- 5} < 2

9 k^{2} - 25

3 x - 3 \geq 9

75 x^{2} y - 120 x y

15 - 8 m = - 73

What are the solutions to the equation p^2+11p-12=0 ?
The solutions to the equation p^2+11p-12=0 are p=1,p=-12
Find the zeros of p^2+11p-12=0
The zeros of p^2+11p-12=0 are p=1,p=-12