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

The solutions to the equation (x-3)(1-x)=-3 are x=0,x=4

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

The zeros of (x-3)(1-x)=-3 are x=0,x=4

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

x = 0, x = 4

Solution steps

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

Expand

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

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

Move

3

to the left side

- x^{2} + 4 x = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- 4 + 4}{2 ( - 1 )} : 0

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

The solutions to the quadratic equation are:

x = 0, x = 4

(2 x + 1)^{2} - 128 = 0

2 (a - 4 x) + a x (3 x 2 + 4) = 2 (6 x 3 + 5)

6 x^{2} - 12 x + 390 = 0

x, x^{2} - 12 x - 36 = 0

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

What are the solutions to the equation (x-3)(1-x)=-3 ?
The solutions to the equation (x-3)(1-x)=-3 are x=0,x=4
Find the zeros of (x-3)(1-x)=-3
The zeros of (x-3)(1-x)=-3 are x=0,x=4