What are the solutions to the equation p^2+15p+14=0 ?

The solutions to the equation p^2+15p+14=0 are p=-1,p=-14

Find the zeros of p^2+15p+14=0

The zeros of p^2+15p+14=0 are p=-1,p=-14

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p^{2} + 15 p + 14 = 0

p = - 1, p = - 14

Solution steps

p^{2} + 15 p + 14 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 1 \cdot 14}{2 \cdot 1}

1 5^{2} - 4 \cdot 1 \cdot 14 = 13

p_{1, 2} = \frac{- 15 \pm 13}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 15 + 13}{2 \cdot 1}, p_{2} = \frac{- 15 - 13}{2 \cdot 1}

p = \frac{- 15 + 13}{2 \cdot 1} : - 1

p = \frac{- 15 - 13}{2 \cdot 1} : - 14

The solutions to the quadratic equation are:

p = - 1, p = - 14

a^{2} + 3 a

80 \cdot \frac{π}{180}

∣ 2 x + 1 ∣ \leq 7

p^{3} - q^{3}

3 w \geq - 27

What are the solutions to the equation p^2+15p+14=0 ?
The solutions to the equation p^2+15p+14=0 are p=-1,p=-14
Find the zeros of p^2+15p+14=0
The zeros of p^2+15p+14=0 are p=-1,p=-14