What are the solutions to the equation (x+1)(x-1)=2(x+5) ?

The solutions to the equation (x+1)(x-1)=2(x+5) are x=1+2sqrt(3),x=1-2sqrt(3)

Find the zeros of (x+1)(x-1)=2(x+5)

The zeros of (x+1)(x-1)=2(x+5) are x=1+2sqrt(3),x=1-2sqrt(3)

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(x + 1) (x - 1) = 2 (x + 5)

x = 1 + 23, x = 1 - 23

Decimal Notation

x = 4.46410 \dots, x = - 2.46410 \dots

Solution steps

(x + 1) (x - 1) = 2 (x + 5)

Expand

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

Expand

2 (x + 5) : 2 x + 10

x^{2} - 1 = 2 x + 10

Move

10

to the left side

x^{2} - 11 = 2 x

Move

2 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 11 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 11) = 43

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

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 4 3}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 4 3}{2 \cdot 1}

x = \frac{- ( - 2 ) + 4 3}{2 \cdot 1} : 1 + 23

x = \frac{- ( - 2 ) - 4 3}{2 \cdot 1} : 1 - 23

The solutions to the quadratic equation are:

x = 1 + 23, x = 1 - 23

x, 6 z = 72 - x^{2} - y^{2}

- 2 x^{2} - 7 x - 6

(x - 3) (x - 2) + \frac{x ( x - 3 )}{2} = (x - 2)^{2}

x^{2} - 20 x = 100

b^{2} - 3 b = 0

What are the solutions to the equation (x+1)(x-1)=2(x+5) ?
The solutions to the equation (x+1)(x-1)=2(x+5) are x=1+2sqrt(3),x=1-2sqrt(3)
Find the zeros of (x+1)(x-1)=2(x+5)
The zeros of (x+1)(x-1)=2(x+5) are x=1+2sqrt(3),x=1-2sqrt(3)