What are the solutions to the equation (-4-x)(1-x)+6=0 ?

The solutions to the equation (-4-x)(1-x)+6=0 are x=-1,x=-2

Find the zeros of (-4-x)(1-x)+6=0

The zeros of (-4-x)(1-x)+6=0 are x=-1,x=-2

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(- 4 - x) (1 - x) + 6 = 0

x = - 1, x = - 2

Solution steps

(- 4 - x) (1 - x) + 6 = 0

Expand

(- 4 - x) (1 - x) + 6 : x^{2} + 3 x + 2

x^{2} + 3 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

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

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

Separate the solutions

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

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

x = \frac{- 3 - 1}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = - 1, x = - 2

9 v^{2} = - 30 v - 25

5 m^{2} + 7 = 52

54 = 6 + 6 a^{2}

r^{2} + 4 r + 5

x^{2} + \frac{9}{2} x + \frac{7}{2} = 0

What are the solutions to the equation (-4-x)(1-x)+6=0 ?
The solutions to the equation (-4-x)(1-x)+6=0 are x=-1,x=-2
Find the zeros of (-4-x)(1-x)+6=0
The zeros of (-4-x)(1-x)+6=0 are x=-1,x=-2