What are the solutions to the equation (x+3)^2+6(x+3)=16 ?

The solutions to the equation (x+3)^2+6(x+3)=16 are x=-1,x=-11

Find the zeros of (x+3)^2+6(x+3)=16

The zeros of (x+3)^2+6(x+3)=16 are x=-1,x=-11

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(x + 3)^{2} + 6 (x + 3) = 16

x = - 1, x = - 11

Solution steps

(x + 3)^{2} + 6 (x + 3) = 16

Expand

(x + 3)^{2} + 6 (x + 3) : x^{2} + 12 x + 27

x^{2} + 12 x + 27 = 16

Move

16

to the left side

x^{2} + 12 x + 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 11}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 11 = 10

x_{1, 2} = \frac{- 12 \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 10}{2 \cdot 1}, x_{2} = \frac{- 12 - 10}{2 \cdot 1}

x = \frac{- 12 + 10}{2 \cdot 1} : - 1

x = \frac{- 12 - 10}{2 \cdot 1} : - 11

The solutions to the quadratic equation are:

x = - 1, x = - 11

5 x + x^{2} = 84

(4 x^{2} - 5 x + 13) + (- 11 x^{2} + 5 x + 4) = a x^{2} + 17

1 - x - 6 x^{2} = 0

2 (x - \frac{m}{2}) (x - (m - 1)) = 0

x, z = 3 x^{2} - 2 x y + y

What are the solutions to the equation (x+3)^2+6(x+3)=16 ?
The solutions to the equation (x+3)^2+6(x+3)=16 are x=-1,x=-11
Find the zeros of (x+3)^2+6(x+3)=16
The zeros of (x+3)^2+6(x+3)=16 are x=-1,x=-11