What are the solutions to the equation (x+3)^2+6(x+3)+5=0 ?

The solutions to the equation (x+3)^2+6(x+3)+5=0 are x=-4,x=-8

Find the zeros of (x+3)^2+6(x+3)+5=0

The zeros of (x+3)^2+6(x+3)+5=0 are x=-4,x=-8

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(x + 3)^{2} + 6 (x + 3) + 5 = 0

x = - 4, x = - 8

Solution steps

(x + 3)^{2} + 6 (x + 3) + 5 = 0

Expand

(x + 3)^{2} + 6 (x + 3) + 5 : x^{2} + 12 x + 32

x^{2} + 12 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 32 = 4

x_{1, 2} = \frac{- 12 \pm 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 4}{2 \cdot 1}, x_{2} = \frac{- 12 - 4}{2 \cdot 1}

x = \frac{- 12 + 4}{2 \cdot 1} : - 4

x = \frac{- 12 - 4}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = - 4, x = - 8

- 2 x^{2} - 4 x - 1 = 0

x^{2} - 5 x + 13 = 11

p^{2} - 12 p = 0

2 t^{2} - 3 t - 1 = 0

(x - 2) (x + 5) = 3

What are the solutions to the equation (x+3)^2+6(x+3)+5=0 ?
The solutions to the equation (x+3)^2+6(x+3)+5=0 are x=-4,x=-8
Find the zeros of (x+3)^2+6(x+3)+5=0
The zeros of (x+3)^2+6(x+3)+5=0 are x=-4,x=-8