What are the solutions to the equation (1-x)(-3-x)+5=0 ?

The solutions to the equation (1-x)(-3-x)+5=0 are x=-1+i,x=-1-i

Find the zeros of (1-x)(-3-x)+5=0

The zeros of (1-x)(-3-x)+5=0 are x=-1+i,x=-1-i

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(1 - x) (- 3 - x) + 5 = 0

x = - 1 + i, x = - 1 - i

Solution steps

(1 - x) (- 3 - x) + 5 = 0

Expand

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

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

Solve with the quadratic formula

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

Simplify

2^{2} - 4 \cdot 1 \cdot 2 : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1 + i, x = - 1 - i

x, 9 x^{2} - 4 y^{2} + 18 x - 16 y = 7

4 x^{2} + 34 x - 25 = 0

49 x^{2} - 7 x - \frac{3}{4}

5 x^{2} + 3 x + 7 = 0

(m + 4) x^{2} - 2 m x + 2 = 0

What are the solutions to the equation (1-x)(-3-x)+5=0 ?
The solutions to the equation (1-x)(-3-x)+5=0 are x=-1+i,x=-1-i
Find the zeros of (1-x)(-3-x)+5=0
The zeros of (1-x)(-3-x)+5=0 are x=-1+i,x=-1-i