What are the solutions to the equation (x+5)2+6=2(x+4)(x+2) ?

The solutions to the equation (x+5)2+6=2(x+4)(x+2) are x=0,x=-5

Find the zeros of (x+5)2+6=2(x+4)(x+2)

The zeros of (x+5)2+6=2(x+4)(x+2) are x=0,x=-5

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(x + 5) 2 + 6 = 2 (x + 4) (x + 2)

x = 0, x = - 5

Solution steps

(x + 5) \cdot 2 + 6 = 2 (x + 4) (x + 2)

Expand

(x + 5) \cdot 2 + 6 : 2 x + 16

Expand

2 (x + 4) (x + 2) : 2 x^{2} + 12 x + 16

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

Switch sides

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

Move

16

to the left side

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

Move

2 x

to the left side

2 x^{2} + 10 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

1 0^{2} - 4 \cdot 2 \cdot 0 = 10

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

Separate the solutions

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

x = \frac{- 10 + 10}{2 \cdot 2} : 0

x = \frac{- 10 - 10}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

x = 0, x = - 5

16 x^{2} = 11

4 x^{2} - 8 x + 7

4 x^{2} - 8 x + 2

0.01 y^{2} - 0.02 y - 0.01 = 0

3 v^{2} - 48 = 0

What are the solutions to the equation (x+5)2+6=2(x+4)(x+2) ?
The solutions to the equation (x+5)2+6=2(x+4)(x+2) are x=0,x=-5
Find the zeros of (x+5)2+6=2(x+4)(x+2)
The zeros of (x+5)2+6=2(x+4)(x+2) are x=0,x=-5