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

The solutions to the equation (x+1)(x-3)=3 are x=1+sqrt(7),x=1-sqrt(7)

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

The zeros of (x+1)(x-3)=3 are x=1+sqrt(7),x=1-sqrt(7)

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

x = 1 + 7, x = 1 - 7

Decimal Notation

x = 3.64575 \dots, x = - 1.64575 \dots

Solution steps

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

Expand

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

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

Move

3

to the left side

x^{2} - 2 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 6) = 27

x_{1, 2} = \frac{- ( - 2 ) \pm 2 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 7}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 7}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 7}{2 \cdot 1} : 1 + 7

x = \frac{- ( - 2 ) - 2 7}{2 \cdot 1} : 1 - 7

The solutions to the quadratic equation are:

x = 1 + 7, x = 1 - 7

x, x^{2} - y^{2} = z

2 X^{2} + 12 X + 16 = 0

x^{2} + 5 x + 8 = 4

55 w = - 20 w^{(2)} - 30

x^{2} - 11 x - 17 = 0

What are the solutions to the equation (x+1)(x-3)=3 ?
The solutions to the equation (x+1)(x-3)=3 are x=1+sqrt(7),x=1-sqrt(7)
Find the zeros of (x+1)(x-3)=3
The zeros of (x+1)(x-3)=3 are x=1+sqrt(7),x=1-sqrt(7)