What are the solutions to the equation (1-x)(-3-x)-2=0 ?

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

Find the zeros of (1-x)(-3-x)-2=0

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

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(1 - x) (- 3 - x) - 2 = 0

x = - 1 + 6, x = - 1 - 6

Decimal Notation

x = 1.44948 \dots, x = - 3.44948 \dots

Solution steps

(1 - x) (- 3 - x) - 2 = 0

Expand

(1 - x) (- 3 - x) - 2 : x^{2} + 2 x - 5

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 5) = 26

x_{1, 2} = \frac{- 2 \pm 2 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 2 6}{2 \cdot 1}, x_{2} = \frac{- 2 - 2 6}{2 \cdot 1}

x = \frac{- 2 + 2 6}{2 \cdot 1} : - 1 + 6

x = \frac{- 2 - 2 6}{2 \cdot 1} : - 1 - 6

The solutions to the quadratic equation are:

x = - 1 + 6, x = - 1 - 6

- 2 x^{2} - 16 x - 35 = 0

(2 t - 1) t = 66

(x + 2)^{2} = 2 (x + 6)

t^{2} - 2 t + 0.4082 = 0

\frac{5}{7} x^{2} - x = \frac{2}{7}

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