What are the solutions to the equation 3x-x^2=1-x ?

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

Find the zeros of 3x-x^2=1-x

The zeros of 3x-x^2=1-x are x=2-sqrt(3),x=2+sqrt(3)

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

x = 2 - 3, x = 2 + 3

Decimal Notation

x = 0.26794 \dots, x = 3.73205 \dots

Solution steps

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

Move

x

to the left side

- x^{2} + 4 x = 1

Move

1

to the left side

- x^{2} + 4 x - 1 = 0

Solve with the quadratic formula

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

4^{2} - 4 (- 1) (- 1) = 23

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

Separate the solutions

x_{1} = \frac{- 4 + 2 3}{2 ( - 1 )}, x_{2} = \frac{- 4 - 2 3}{2 ( - 1 )}

x = \frac{- 4 + 2 3}{2 ( - 1 )} : 2 - 3

x = \frac{- 4 - 2 3}{2 ( - 1 )} : 2 + 3

The solutions to the quadratic equation are:

x = 2 - 3, x = 2 + 3

3 x^{2} + 18 x + 252 = 0

x^{2} + 2 x = \frac{1}{2}

d^{2} - 10 d + 5 = 0

2 x + 66 + x = 3 x + 2 + 2 x (4 x)

7 x^{2} - 7 = 0

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