What are the solutions to the equation (x+2)(x+4)=3 ?

The solutions to the equation (x+2)(x+4)=3 are x=-1,x=-5

Find the zeros of (x+2)(x+4)=3

The zeros of (x+2)(x+4)=3 are x=-1,x=-5

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(x + 2) (x + 4) = 3

x = - 1, x = - 5

Solution steps

(x + 2) (x + 4) = 3

Expand

(x + 2) (x + 4) : x^{2} + 6 x + 8

x^{2} + 6 x + 8 = 3

Move

3

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

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

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

Separate the solutions

x_{1} = \frac{- 6 + 4}{2 \cdot 1}, x_{2} = \frac{- 6 - 4}{2 \cdot 1}

x = \frac{- 6 + 4}{2 \cdot 1} : - 1

x = \frac{- 6 - 4}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = - 1, x = - 5

x^{2} - 11 x + 30 = 0

4 r^{2} - 20 r + 25 = 0

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

3 x^{2} - 2 x - 65 = 0

3 x^{2} - 27 x + 75 = 15

What are the solutions to the equation (x+2)(x+4)=3 ?
The solutions to the equation (x+2)(x+4)=3 are x=-1,x=-5
Find the zeros of (x+2)(x+4)=3
The zeros of (x+2)(x+4)=3 are x=-1,x=-5