What are the solutions to the equation (8x-1)(2x+1)=3x ?

The solutions to the equation (8x-1)(2x+1)=3x are x=(-3+sqrt(73))/(32),x=(-3-sqrt(73))/(32)

Find the zeros of (8x-1)(2x+1)=3x

The zeros of (8x-1)(2x+1)=3x are x=(-3+sqrt(73))/(32),x=(-3-sqrt(73))/(32)

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(8 x - 1) (2 x + 1) = 3 x

x = \frac{- 3 + 73}{32}, x = \frac{- 3 - 73}{32}

Decimal Notation

x = 0.17325 \dots, x = - 0.36075 \dots

Solution steps

(8 x - 1) (2 x + 1) = 3 x

Expand

(8 x - 1) (2 x + 1) : 16 x^{2} + 6 x - 1

16 x^{2} + 6 x - 1 = 3 x

Move

3 x

to the left side

16 x^{2} + 3 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 16 ( - 1 )}{2 \cdot 16}

3^{2} - 4 \cdot 16 (- 1) = 73

x_{1, 2} = \frac{- 3 \pm 73}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- 3 + 73}{2 \cdot 16}, x_{2} = \frac{- 3 - 73}{2 \cdot 16}

x = \frac{- 3 + 73}{2 \cdot 16} : \frac{- 3 + 73}{32}

x = \frac{- 3 - 73}{2 \cdot 16} : \frac{- 3 - 73}{32}

The solutions to the quadratic equation are:

x = \frac{- 3 + 73}{32}, x = \frac{- 3 - 73}{32}

\frac{2 ! 4 !}{3 !}

- 2 x < 10

4 a^{2} - 16 a = - 15

\frac{5}{3} + 1

What are the solutions to the equation (8x-1)(2x+1)=3x ?
The solutions to the equation (8x-1)(2x+1)=3x are x=(-3+sqrt(73))/(32),x=(-3-sqrt(73))/(32)
Find the zeros of (8x-1)(2x+1)=3x
The zeros of (8x-1)(2x+1)=3x are x=(-3+sqrt(73))/(32),x=(-3-sqrt(73))/(32)