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

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

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

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

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(x + 5) (x + 2) + 2 = 0

x = - 3, x = - 4

Solution steps

(x + 5) (x + 2) + 2 = 0

Expand

(x + 5) (x + 2) + 2 : x^{2} + 7 x + 12

x^{2} + 7 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot 12 = 1

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

Separate the solutions

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

x = \frac{- 7 + 1}{2 \cdot 1} : - 3

x = \frac{- 7 - 1}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = - 3, x = - 4

x^{2} + 6 x + 8 = 0

2 x^{2} - 12 = 2 x

x, x^{2} y = 4

n^{2} + 10 n + 21 = 0

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

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