What are the solutions to the equation -3x-3=(x-2)(x+1) ?

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

Find the zeros of -3x-3=(x-2)(x+1)

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

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- 3 x - 3 = (x - 2) (x + 1)

x = - 1

Solution steps

- 3 x - 3 = (x - 2) (x + 1)

Expand

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

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

Switch sides

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

Move

3

to the left side

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

Move

3 x

to the left side

x^{2} + 2 x + 1 = 0

Solve with the quadratic formula

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

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

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

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

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

x = - 1

The solution to the quadratic equation is:

x = - 1

x^{2} - 4 x - 14 = - 2

\frac{5 x - 1}{3} > 5

9 - 4 u^{2}

- \frac{1}{3} h \leq 6

9 x^{2} + 5 x = 1

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