What are the solutions to the equation (x-3)(1-x)=1 ?

The solutions to the equation (x-3)(1-x)=1 are x=2

Find the zeros of (x-3)(1-x)=1

The zeros of (x-3)(1-x)=1 are x=2

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(x - 3) (1 - x) = 1

x = 2

Solution steps

(x - 3) (1 - x) = 1

Expand

(x - 3) (1 - x) : 4 x - x^{2} - 3

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

Move

1

to the left side

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

Solve with the quadratic formula

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

4^{2} - 4 (- 1) (- 4) = 0

x_{1, 2} = \frac{- 4 \pm 0}{2 ( - 1 )}

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

\frac{- 4}{2 ( - 1 )} = 2

x = 2

The solution to the quadratic equation is:

x = 2

2 x + 7 \leq 12

(x + 9) (x - 4) < 0

7 (x - 5) - 2 x (x - 5)

1 - \frac{5}{6}

5 = \frac{2 x}{7} + \frac{4}{7} - 3

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