What are the solutions to the equation (4x-1)(2x+3)=18x-4 ?

The solutions to the equation (4x-1)(2x+3)=18x-4 are x=(2+sqrt(2))/4 ,x=(2-sqrt(2))/4

Find the zeros of (4x-1)(2x+3)=18x-4

The zeros of (4x-1)(2x+3)=18x-4 are x=(2+sqrt(2))/4 ,x=(2-sqrt(2))/4

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(4 x - 1) (2 x + 3) = 18 x - 4

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

Solution steps

(4 x - 1) (2 x + 3) = 18 x - 4

Expand

(4 x - 1) (2 x + 3) : 8 x^{2} + 10 x - 3

8 x^{2} + 10 x - 3 = 18 x - 4

Move

4

to the left side

8 x^{2} + 10 x + 1 = 18 x

Move

18 x

to the left side

8 x^{2} - 8 x + 1 = 0

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 8 \cdot 1 = 42

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

Separate the solutions

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

x = \frac{- ( - 8 ) + 4 2}{2 \cdot 8} : \frac{2 + 2}{4}

x = \frac{- ( - 8 ) - 4 2}{2 \cdot 8} : \frac{2 - 2}{4}

The solutions to the quadratic equation are:

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

x^{2} + \frac{8}{3} x + \frac{5}{3} = 0

\frac{( 2 ) ^{2} - 4}{( 2 ) ^{2} + ( 2 ) - 6}

2 (\frac{45}{53}) (\frac{28}{53})

- 9 < \frac{x}{5} - 2

2 π r h + 2 π r^{2}

What are the solutions to the equation (4x-1)(2x+3)=18x-4 ?
The solutions to the equation (4x-1)(2x+3)=18x-4 are x=(2+sqrt(2))/4 ,x=(2-sqrt(2))/4
Find the zeros of (4x-1)(2x+3)=18x-4
The zeros of (4x-1)(2x+3)=18x-4 are x=(2+sqrt(2))/4 ,x=(2-sqrt(2))/4