What are the solutions to the equation p^2+11p+28=0 ?

The solutions to the equation p^2+11p+28=0 are p=-4,p=-7

Find the zeros of p^2+11p+28=0

The zeros of p^2+11p+28=0 are p=-4,p=-7

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p^{2} + 11 p + 28 = 0

p = - 4, p = - 7

Solution steps

p^{2} + 11 p + 28 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

1 1^{2} - 4 \cdot 1 \cdot 28 = 3

p_{1, 2} = \frac{- 11 \pm 3}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 11 + 3}{2 \cdot 1}, p_{2} = \frac{- 11 - 3}{2 \cdot 1}

p = \frac{- 11 + 3}{2 \cdot 1} : - 4

p = \frac{- 11 - 3}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

p = - 4, p = - 7

3 x (x - 2) - 2 (x - 1) (x + 3) = 0

lo g_{b} (7) x^{2} - lo g_{b} (49) = 0

2 x^{2} = 5 x - 3

5 x^{2} + 7 x - 10 = 0

(y + 11)^{2} = 121

What are the solutions to the equation p^2+11p+28=0 ?
The solutions to the equation p^2+11p+28=0 are p=-4,p=-7
Find the zeros of p^2+11p+28=0
The zeros of p^2+11p+28=0 are p=-4,p=-7