What are the solutions to the equation 2x(4x-1)=9 ?

The solutions to the equation 2x(4x-1)=9 are x=(1+sqrt(73))/8 ,x=(1-sqrt(73))/8

Find the zeros of 2x(4x-1)=9

The zeros of 2x(4x-1)=9 are x=(1+sqrt(73))/8 ,x=(1-sqrt(73))/8

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2 x (4 x - 1) = 9

x = \frac{1 + 73}{8}, x = \frac{1 - 73}{8}

Decimal Notation

x = 1.19300 \dots, x = - 0.94300 \dots

Solution steps

2 x (4 x - 1) = 9

Expand

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

8 x^{2} - 2 x = 9

Move

9

to the left side

8 x^{2} - 2 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 9 )}{2 \cdot 8}

(- 2)^{2} - 4 \cdot 8 (- 9) = 273

x_{1, 2} = \frac{- ( - 2 ) \pm 2 73}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 73}{2 \cdot 8}, x_{2} = \frac{- ( - 2 ) - 2 73}{2 \cdot 8}

x = \frac{- ( - 2 ) + 2 73}{2 \cdot 8} : \frac{1 + 73}{8}

x = \frac{- ( - 2 ) - 2 73}{2 \cdot 8} : \frac{1 - 73}{8}

The solutions to the quadratic equation are:

x = \frac{1 + 73}{8}, x = \frac{1 - 73}{8}

- \frac{4}{1}

\frac{1}{10} \times \frac{9}{10} + \frac{9}{10} \times \frac{1}{10}

- 2 + 3 (\frac{5}{6}) \geq \frac{1}{4}

e^{(\frac{iπ}{6})}

4 x^{3} - 10 x

What are the solutions to the equation 2x(4x-1)=9 ?
The solutions to the equation 2x(4x-1)=9 are x=(1+sqrt(73))/8 ,x=(1-sqrt(73))/8
Find the zeros of 2x(4x-1)=9
The zeros of 2x(4x-1)=9 are x=(1+sqrt(73))/8 ,x=(1-sqrt(73))/8