What are the solutions to the equation 9x-2-2x(3x-1)=4x^2 ?

The solutions to the equation 9x-2-2x(3x-1)=4x^2 are x=(11-sqrt(41))/(20),x=(11+sqrt(41))/(20)

Find the zeros of 9x-2-2x(3x-1)=4x^2

The zeros of 9x-2-2x(3x-1)=4x^2 are x=(11-sqrt(41))/(20),x=(11+sqrt(41))/(20)

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9 x - 2 - 2 x (3 x - 1) = 4 x^{2}

x = \frac{11 - 41}{20}, x = \frac{11 + 41}{20}

Decimal Notation

x = 0.22984 \dots, x = 0.87015 \dots

Solution steps

9 x - 2 - 2 x (3 x - 1) = 4 x^{2}

Expand

9 x - 2 - 2 x (3 x - 1) : - 6 x^{2} + 11 x - 2

- 6 x^{2} + 11 x - 2 = 4 x^{2}

Move

4 x^{2}

to the left side

- 10 x^{2} + 11 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 ( - 10 ) ( - 2 )}{2 ( - 10 )}

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

x_{1, 2} = \frac{- 11 \pm 41}{2 ( - 10 )}

Separate the solutions

x_{1} = \frac{- 11 + 41}{2 ( - 10 )}, x_{2} = \frac{- 11 - 41}{2 ( - 10 )}

x = \frac{- 11 + 41}{2 ( - 10 )} : \frac{11 - 41}{20}

x = \frac{- 11 - 41}{2 ( - 10 )} : \frac{11 + 41}{20}

The solutions to the quadratic equation are:

x = \frac{11 - 41}{20}, x = \frac{11 + 41}{20}

5 y^{2} = 7 y + 6

2 s^{2} + 15 s + 40 = 0

x, x^{2} + 10 x + 24 = 0

a^{2} - a - 2 = 0

x^{2} + 18 x = 9

What are the solutions to the equation 9x-2-2x(3x-1)=4x^2 ?
The solutions to the equation 9x-2-2x(3x-1)=4x^2 are x=(11-sqrt(41))/(20),x=(11+sqrt(41))/(20)
Find the zeros of 9x-2-2x(3x-1)=4x^2
The zeros of 9x-2-2x(3x-1)=4x^2 are x=(11-sqrt(41))/(20),x=(11+sqrt(41))/(20)