What are the solutions to the equation (x-1)(x+1)=-(4x+5) ?

The solutions to the equation (x-1)(x+1)=-(4x+5) are x=-2

Find the zeros of (x-1)(x+1)=-(4x+5)

The zeros of (x-1)(x+1)=-(4x+5) are x=-2

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(x - 1) (x + 1) = - (4 x + 5)

x = - 2

Solution steps

(x - 1) (x + 1) = - (4 x + 5)

Expand

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

Expand

- (4 x + 5) : - 4 x - 5

x^{2} - 1 = - 4 x - 5

Move

5

to the left side

x^{2} + 4 = - 4 x

Move

4 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot 4 = 0

x_{1, 2} = \frac{- 4 \pm 0}{2 \cdot 1}

x = \frac{- 4}{2 \cdot 1}

\frac{- 4}{2 \cdot 1} = - 2

x = - 2

The solution to the quadratic equation is:

x = - 2

76 x^{2} - 140 x - 846 = 0

9.81 \cdot 450 \cdot 0.05091 = 36.2925 x^{2}

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

27 x^{2} + 19 x - 37 = 0

- 4 w^{2} + 16 w - 16 = 0

What are the solutions to the equation (x-1)(x+1)=-(4x+5) ?
The solutions to the equation (x-1)(x+1)=-(4x+5) are x=-2
Find the zeros of (x-1)(x+1)=-(4x+5)
The zeros of (x-1)(x+1)=-(4x+5) are x=-2