What are the solutions to the equation (5x-1)(2x+1)=2x ?

The solutions to the equation (5x-1)(2x+1)=2x are x=(-1+sqrt(41))/(20),x=(-1-sqrt(41))/(20)

Find the zeros of (5x-1)(2x+1)=2x

The zeros of (5x-1)(2x+1)=2x are x=(-1+sqrt(41))/(20),x=(-1-sqrt(41))/(20)

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(5 x - 1) (2 x + 1) = 2 x

x = \frac{- 1 + 41}{20}, x = \frac{- 1 - 41}{20}

Decimal Notation

x = 0.27015 \dots, x = - 0.37015 \dots

Solution steps

(5 x - 1) (2 x + 1) = 2 x

Expand

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

10 x^{2} + 3 x - 1 = 2 x

Move

2 x

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 10 ( - 1 )}{2 \cdot 10}

1^{2} - 4 \cdot 10 (- 1) = 41

x_{1, 2} = \frac{- 1 \pm 41}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- 1 + 41}{2 \cdot 10}, x_{2} = \frac{- 1 - 41}{2 \cdot 10}

x = \frac{- 1 + 41}{2 \cdot 10} : \frac{- 1 + 41}{20}

x = \frac{- 1 - 41}{2 \cdot 10} : \frac{- 1 - 41}{20}

The solutions to the quadratic equation are:

x = \frac{- 1 + 41}{20}, x = \frac{- 1 - 41}{20}

- 5 \times 7 + 9 \times \frac{5}{5}

117 x^{2} + 77 x - 24

∣ 2 x + 6 ∣ - 15 < - 1

(x - 5) (x - 2)^{2} < 0

2 x^{2} - 72 x^{2} = 0

What are the solutions to the equation (5x-1)(2x+1)=2x ?
The solutions to the equation (5x-1)(2x+1)=2x are x=(-1+sqrt(41))/(20),x=(-1-sqrt(41))/(20)
Find the zeros of (5x-1)(2x+1)=2x
The zeros of (5x-1)(2x+1)=2x are x=(-1+sqrt(41))/(20),x=(-1-sqrt(41))/(20)