What are the solutions to the equation (3-x)(-2-x)+6=0 ?

The solutions to the equation (3-x)(-2-x)+6=0 are x=1,x=0

Find the zeros of (3-x)(-2-x)+6=0

The zeros of (3-x)(-2-x)+6=0 are x=1,x=0

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(3 - x) (- 2 - x) + 6 = 0

x = 1, x = 0

Solution steps

(3 - x) (- 2 - x) + 6 = 0

Expand

(3 - x) (- 2 - x) + 6 : x^{2} - x

x^{2} - x = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1, x = 0

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

6 r^{2} - 4 r - 80 = 0

12 x^{2} - 128 x + 252 = 0

y, 5 x^{2} + 2 x^{2} y + y^{2} = 8

x^{2} = 8 x + 10

What are the solutions to the equation (3-x)(-2-x)+6=0 ?
The solutions to the equation (3-x)(-2-x)+6=0 are x=1,x=0
Find the zeros of (3-x)(-2-x)+6=0
The zeros of (3-x)(-2-x)+6=0 are x=1,x=0