What are the solutions to the equation 2(x+2)(x-4)+16=4(3-x) ?

The solutions to the equation 2(x+2)(x-4)+16=4(3-x) are x=sqrt(6),x=-sqrt(6)

Find the zeros of 2(x+2)(x-4)+16=4(3-x)

The zeros of 2(x+2)(x-4)+16=4(3-x) are x=sqrt(6),x=-sqrt(6)

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2 (x + 2) (x - 4) + 16 = 4 (3 - x)

x = 6, x = - 6

Decimal Notation

x = 2.44948 \dots, x = - 2.44948 \dots

Solution steps

2 (x + 2) (x - 4) + 16 = 4 (3 - x)

Expand

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

Expand

4 (3 - x) : 12 - 4 x

2 x^{2} - 4 x = 12 - 4 x

Move

4 x

to the left side

2 x^{2} = 12

Divide both sides by

2

x^{2} = 6

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 6, x = - 6

x^{2} + 14.14 x + 100 = 0

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

x^{2} + 10 x - 10 = 0

2 x^{2} + 10 x = 0

1 1^{2} + 8^{2} = x^{2}

What are the solutions to the equation 2(x+2)(x-4)+16=4(3-x) ?
The solutions to the equation 2(x+2)(x-4)+16=4(3-x) are x=sqrt(6),x=-sqrt(6)
Find the zeros of 2(x+2)(x-4)+16=4(3-x)
The zeros of 2(x+2)(x-4)+16=4(3-x) are x=sqrt(6),x=-sqrt(6)