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

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

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

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

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

x = - 2, x = 1

Solution steps

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

Move

1

to the left side

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

Move

x

to the left side

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 (- 1) \cdot 2 = 3

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 2, x = 1

x, y (x - 3)^{2} = x^{2} + 9

34 \cdot x^{2} + 68 \cdot x - 510 = 0

- 16 t^{2} + 116 t + 101 = 0

(12 y - 2 (\frac{18 - 3 y}{4}) y - 3 y^{2}) = 0

(3 x - 1) (x - 2) = 22

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