What are the solutions to the equation 8x+2=(2x+4)4x+5 ?

The solutions to the equation 8x+2=(2x+4)4x+5 are x=-1/2+i(sqrt(2))/4 ,x=-1/2-i(sqrt(2))/4

Find the zeros of 8x+2=(2x+4)4x+5

The zeros of 8x+2=(2x+4)4x+5 are x=-1/2+i(sqrt(2))/4 ,x=-1/2-i(sqrt(2))/4

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

x = - \frac{1}{2} + i \frac{2}{4}, x = - \frac{1}{2} - i \frac{2}{4}

Solution steps

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

Expand

(2 x + 4) \cdot 4 x + 5 : 8 x^{2} + 16 x + 5

8 x + 2 = 8 x^{2} + 16 x + 5

Switch sides

8 x^{2} + 16 x + 5 = 8 x + 2

Move

2

to the left side

8 x^{2} + 16 x + 3 = 8 x

Move

8 x

to the left side

8 x^{2} + 8 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

Simplify

8^{2} - 4 \cdot 8 \cdot 3 : 42 i

x_{1, 2} = \frac{- 8 \pm 4 2 i}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 8 + 4 2 i}{2 \cdot 8}, x_{2} = \frac{- 8 - 4 2 i}{2 \cdot 8}

x = \frac{- 8 + 4 2 i}{2 \cdot 8} : - \frac{1}{2} + i \frac{2}{4}

x = \frac{- 8 - 4 2 i}{2 \cdot 8} : - \frac{1}{2} - i \frac{2}{4}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{2}{4}, x = - \frac{1}{2} - i \frac{2}{4}

\frac{11}{9}

- 10 \cdot (- 9 + 8 i)

\frac{1}{1.8 \cdot 1 0 ^{- 5}}

2 + \frac{x - 1}{2} = x - 1

16 x^{2} - 81 y^{2}

What are the solutions to the equation 8x+2=(2x+4)4x+5 ?
The solutions to the equation 8x+2=(2x+4)4x+5 are x=-1/2+i(sqrt(2))/4 ,x=-1/2-i(sqrt(2))/4
Find the zeros of 8x+2=(2x+4)4x+5
The zeros of 8x+2=(2x+4)4x+5 are x=-1/2+i(sqrt(2))/4 ,x=-1/2-i(sqrt(2))/4