What are the solutions to the equation (x+2)^2+4(x+2)-5=0 ?

The solutions to the equation (x+2)^2+4(x+2)-5=0 are x=-1,x=-7

Find the zeros of (x+2)^2+4(x+2)-5=0

The zeros of (x+2)^2+4(x+2)-5=0 are x=-1,x=-7

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(x + 2)^{2} + 4 (x + 2) - 5 = 0

x = - 1, x = - 7

Solution steps

(x + 2)^{2} + 4 (x + 2) - 5 = 0

Expand

(x + 2)^{2} + 4 (x + 2) - 5 : x^{2} + 8 x + 7

x^{2} + 8 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 7}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot 7 = 6

x_{1, 2} = \frac{- 8 \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 8 + 6}{2 \cdot 1}, x_{2} = \frac{- 8 - 6}{2 \cdot 1}

x = \frac{- 8 + 6}{2 \cdot 1} : - 1

x = \frac{- 8 - 6}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

x = - 1, x = - 7

3 x^{2} - 6 x + 5 = 0

5 x^{2} + 4 x - 1

x^{3} + 3 x^{2} - 4 x

6 x^{2} + 5 x - 6 = 0

\frac{1}{5} - 2

What are the solutions to the equation (x+2)^2+4(x+2)-5=0 ?
The solutions to the equation (x+2)^2+4(x+2)-5=0 are x=-1,x=-7
Find the zeros of (x+2)^2+4(x+2)-5=0
The zeros of (x+2)^2+4(x+2)-5=0 are x=-1,x=-7