What are the solutions to the equation (p+1)^2+5(p+1)=0 ?

The solutions to the equation (p+1)^2+5(p+1)=0 are p=-1,p=-6

Find the zeros of (p+1)^2+5(p+1)=0

The zeros of (p+1)^2+5(p+1)=0 are p=-1,p=-6

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(p + 1)^{2} + 5 (p + 1) = 0

p = - 1, p = - 6

Solution steps

(p + 1)^{2} + 5 (p + 1) = 0

Expand

(p + 1)^{2} + 5 (p + 1) : p^{2} + 7 p + 6

p^{2} + 7 p + 6 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 6 = 5

p_{1, 2} = \frac{- 7 \pm 5}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 7 + 5}{2 \cdot 1}, p_{2} = \frac{- 7 - 5}{2 \cdot 1}

p = \frac{- 7 + 5}{2 \cdot 1} : - 1

p = \frac{- 7 - 5}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

p = - 1, p = - 6

2 r \leq - 8

4 x - 3 = 5 x + 6

\frac{( 9 x - 11 ) ( 2 x + 7 )}{( 3 x - 8 ) ^{3}} > 0

∣ 7 x + 28 ∣ = 0

s^{2} + 12 s - 13

What are the solutions to the equation (p+1)^2+5(p+1)=0 ?
The solutions to the equation (p+1)^2+5(p+1)=0 are p=-1,p=-6
Find the zeros of (p+1)^2+5(p+1)=0
The zeros of (p+1)^2+5(p+1)=0 are p=-1,p=-6