What are the solutions to the equation (4-x)(1-x)+2=0 ?

The solutions to the equation (4-x)(1-x)+2=0 are x=3,x=2

Find the zeros of (4-x)(1-x)+2=0

The zeros of (4-x)(1-x)+2=0 are x=3,x=2

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(4 - x) (1 - x) + 2 = 0

x = 3, x = 2

Solution steps

(4 - x) (1 - x) + 2 = 0

Expand

(4 - x) (1 - x) + 2 : x^{2} - 5 x + 6

x^{2} - 5 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 6 = 1

x_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 1}

x = \frac{- ( - 5 ) + 1}{2 \cdot 1} : 3

x = \frac{- ( - 5 ) - 1}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 3, x = 2

2 z^{2} + 2.5 z + 3.75 = 0

6 x^{2} = 29 x + 5

2 x (x - 3) = x^{2} - 9

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

4 x^{2} + x + 12 = 0

What are the solutions to the equation (4-x)(1-x)+2=0 ?
The solutions to the equation (4-x)(1-x)+2=0 are x=3,x=2
Find the zeros of (4-x)(1-x)+2=0
The zeros of (4-x)(1-x)+2=0 are x=3,x=2