What are the solutions to the equation (4x+1)(8x-1)=16x+3 ?

The solutions to the equation (4x+1)(8x-1)=16x+3 are x=(3+sqrt(41))/(16),x=(3-sqrt(41))/(16)

Find the zeros of (4x+1)(8x-1)=16x+3

The zeros of (4x+1)(8x-1)=16x+3 are x=(3+sqrt(41))/(16),x=(3-sqrt(41))/(16)

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(4 x + 1) (8 x - 1) = 16 x + 3

x = \frac{3 + 41}{16}, x = \frac{3 - 41}{16}

Decimal Notation

x = 0.58769 \dots, x = - 0.21269 \dots

Solution steps

(4 x + 1) (8 x - 1) = 16 x + 3

Expand

(4 x + 1) (8 x - 1) : 32 x^{2} + 4 x - 1

32 x^{2} + 4 x - 1 = 16 x + 3

Move

3

to the left side

32 x^{2} + 4 x - 4 = 16 x

Move

16 x

to the left side

32 x^{2} - 12 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 32 ( - 4 )}{2 \cdot 32}

(- 12)^{2} - 4 \cdot 32 (- 4) = 441

x_{1, 2} = \frac{- ( - 12 ) \pm 4 41}{2 \cdot 32}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 4 41}{2 \cdot 32}, x_{2} = \frac{- ( - 12 ) - 4 41}{2 \cdot 32}

x = \frac{- ( - 12 ) + 4 41}{2 \cdot 32} : \frac{3 + 41}{16}

x = \frac{- ( - 12 ) - 4 41}{2 \cdot 32} : \frac{3 - 41}{16}

The solutions to the quadratic equation are:

x = \frac{3 + 41}{16}, x = \frac{3 - 41}{16}

27 x^{2} = 12

5 y^{2} + 4 y - 2 = 0

- 2 x^{2} + 6 x + 7 = 2 x - 3

x^{2} + 2 x + 20 = - 9

15 p^{2} + 80 p + 80 = 0

What are the solutions to the equation (4x+1)(8x-1)=16x+3 ?
The solutions to the equation (4x+1)(8x-1)=16x+3 are x=(3+sqrt(41))/(16),x=(3-sqrt(41))/(16)
Find the zeros of (4x+1)(8x-1)=16x+3
The zeros of (4x+1)(8x-1)=16x+3 are x=(3+sqrt(41))/(16),x=(3-sqrt(41))/(16)