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

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

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

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

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

x = 1 + 5, x = 1 - 5

Decimal Notation

x = 3.23606 \dots, x = - 1.23606 \dots

Solution steps

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

Expand

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

x^{2} - 2 x - 3 = 1

Move

1

to the left side

x^{2} - 2 x - 4 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 4) = 25

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + 5, x = 1 - 5

- 7 x + 11 = - 2 x^{2}

3 x^{2} - 35 x + 50 = 0

2 z^{2} - z - 2 = 0

x^{2} - 12 x - 30 = 0

x + 3 = x^{2} + x

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