What are the solutions to the equation 6x-2x^2=2x-3 ?

The solutions to the equation 6x-2x^2=2x-3 are x=-(-2+sqrt(10))/2 ,x=(2+sqrt(10))/2

Find the zeros of 6x-2x^2=2x-3

The zeros of 6x-2x^2=2x-3 are x=-(-2+sqrt(10))/2 ,x=(2+sqrt(10))/2

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6 x - 2 x^{2} = 2 x - 3

x = - \frac{- 2 + 10}{2}, x = \frac{2 + 10}{2}

Decimal Notation

x = - 0.58113 \dots, x = 2.58113 \dots

Solution steps

6 x - 2 x^{2} = 2 x - 3

Move

3

to the left side

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

Move

2 x

to the left side

- 2 x^{2} + 4 x + 3 = 0

Solve with the quadratic formula

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

4^{2} - 4 (- 2) \cdot 3 = 210

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

Separate the solutions

x_{1} = \frac{- 4 + 2 10}{2 ( - 2 )}, x_{2} = \frac{- 4 - 2 10}{2 ( - 2 )}

x = \frac{- 4 + 2 10}{2 ( - 2 )} : - \frac{- 2 + 10}{2}

x = \frac{- 4 - 2 10}{2 ( - 2 )} : \frac{2 + 10}{2}

The solutions to the quadratic equation are:

x = - \frac{- 2 + 10}{2}, x = \frac{2 + 10}{2}

2 x^{2} - 30 = 2

x (x - 8) = 9

x, - x^{2} + \frac{36 y ^{2}}{25} = 0

x^{2} - 4 x + 4 = 0

t^{2} = 81

What are the solutions to the equation 6x-2x^2=2x-3 ?
The solutions to the equation 6x-2x^2=2x-3 are x=-(-2+sqrt(10))/2 ,x=(2+sqrt(10))/2
Find the zeros of 6x-2x^2=2x-3
The zeros of 6x-2x^2=2x-3 are x=-(-2+sqrt(10))/2 ,x=(2+sqrt(10))/2