What are the solutions to the equation (x-2)(x+1)=6 ?

The solutions to the equation (x-2)(x+1)=6 are x=(1+sqrt(33))/2 ,x=(1-sqrt(33))/2

Find the zeros of (x-2)(x+1)=6

The zeros of (x-2)(x+1)=6 are x=(1+sqrt(33))/2 ,x=(1-sqrt(33))/2

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(x - 2) (x + 1) = 6

x = \frac{1 + 33}{2}, x = \frac{1 - 33}{2}

Decimal Notation

x = 3.37228 \dots, x = - 2.37228 \dots

Solution steps

(x - 2) (x + 1) = 6

Expand

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

x^{2} - x - 2 = 6

Move

6

to the left side

x^{2} - x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 8) = 33

x_{1, 2} = \frac{- ( - 1 ) \pm 33}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 33}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 33}{2 \cdot 1}

x = \frac{- ( - 1 ) + 33}{2 \cdot 1} : \frac{1 + 33}{2}

x = \frac{- ( - 1 ) - 33}{2 \cdot 1} : \frac{1 - 33}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 33}{2}, x = \frac{1 - 33}{2}

1 2^{2} + x^{2} = 1 6^{2}

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

(8 x 3 + 125) = (2 x + a) (b x 2 - 10 x + c)

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

x^{2} + 12 = x

What are the solutions to the equation (x-2)(x+1)=6 ?
The solutions to the equation (x-2)(x+1)=6 are x=(1+sqrt(33))/2 ,x=(1-sqrt(33))/2
Find the zeros of (x-2)(x+1)=6
The zeros of (x-2)(x+1)=6 are x=(1+sqrt(33))/2 ,x=(1-sqrt(33))/2