What are the solutions to the equation (x+4)^2-(x+4)-6=0 ?

The solutions to the equation (x+4)^2-(x+4)-6=0 are x=-1,x=-6

Find the zeros of (x+4)^2-(x+4)-6=0

The zeros of (x+4)^2-(x+4)-6=0 are x=-1,x=-6

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(x + 4)^{2} - (x + 4) - 6 = 0

x = - 1, x = - 6

Solution steps

(x + 4)^{2} - (x + 4) - 6 = 0

Expand

(x + 4)^{2} - (x + 4) - 6 : x^{2} + 7 x + 6

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1, x = - 6

q, z (1 - q^{2}) = 2 (a q x + q y)

- x (x - 6) = 0

3 x (x + 4) + 6 (x^{2} - 1) = - 10

c, m \cdot b \cdot a = n \cdot k \cdot c^{2} \cdot s \cdot p + m \cdot c \cdot a

x + 2 = - 3 x^{2}

What are the solutions to the equation (x+4)^2-(x+4)-6=0 ?
The solutions to the equation (x+4)^2-(x+4)-6=0 are x=-1,x=-6
Find the zeros of (x+4)^2-(x+4)-6=0
The zeros of (x+4)^2-(x+4)-6=0 are x=-1,x=-6