What are the solutions to the equation 640000=2(2x)(x)+2(2x)(x)+2(2x)(2x) ?

The solutions to the equation 640000=2(2x)(x)+2(2x)(x)+2(2x)(2x) are x=200,x=-200

Find the zeros of 640000=2(2x)(x)+2(2x)(x)+2(2x)(2x)

The zeros of 640000=2(2x)(x)+2(2x)(x)+2(2x)(2x) are x=200,x=-200

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640000 = 2 (2 x) (x) + 2 (2 x) (x) + 2 (2 x) (2 x)

x = 200, x = - 200

Solution steps

640000 = 2 (2 x) (x) + 2 (2 x) (x) + 2 (2 x) (2 x)

Expand

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

640000 = 16 x^{2}

Switch sides

16 x^{2} = 640000

Move

640000

to the left side

16 x^{2} - 640000 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 16 ( - 640000 )}{2 \cdot 16}

0^{2} - 4 \cdot 16 (- 640000) = 6400

x_{1, 2} = \frac{- 0 \pm 6400}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- 0 + 6400}{2 \cdot 16}, x_{2} = \frac{- 0 - 6400}{2 \cdot 16}

x = \frac{- 0 + 6400}{2 \cdot 16} : 200

x = \frac{- 0 - 6400}{2 \cdot 16} : - 200

The solutions to the quadratic equation are:

x = 200, x = - 200

0 = 3 x^{2} + 4 x - 15

18 a^{2} - 63 a + 10 = 0

8 x^{2} + 10 x = 3

x^{2} + 4 x - 1 = 7

c^{1} + c^{2} = - 2

What are the solutions to the equation 640000=2(2x)(x)+2(2x)(x)+2(2x)(2x) ?
The solutions to the equation 640000=2(2x)(x)+2(2x)(x)+2(2x)(2x) are x=200,x=-200
Find the zeros of 640000=2(2x)(x)+2(2x)(x)+2(2x)(2x)
The zeros of 640000=2(2x)(x)+2(2x)(x)+2(2x)(2x) are x=200,x=-200