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

The solutions to the equation (x-2)(x+1)=5 are x=(1+sqrt(29))/2 ,x=(1-sqrt(29))/2

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

The zeros of (x-2)(x+1)=5 are x=(1+sqrt(29))/2 ,x=(1-sqrt(29))/2

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

x = \frac{1 + 29}{2}, x = \frac{1 - 29}{2}

Decimal Notation

x = 3.19258 \dots, x = - 2.19258 \dots

Solution steps

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

Expand

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

x^{2} - x - 2 = 5

Move

5

to the left side

x^{2} - x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 7) = 29

x_{1, 2} = \frac{- ( - 1 ) \pm 29}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 29}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 29}{2 \cdot 1}

x = \frac{- ( - 1 ) + 29}{2 \cdot 1} : \frac{1 + 29}{2}

x = \frac{- ( - 1 ) - 29}{2 \cdot 1} : \frac{1 - 29}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 29}{2}, x = \frac{1 - 29}{2}

8 - 2 (x - 5) = x - 3

b^{2} - 6 b + 8

2 (2 x + 1) (x - 3) = 0

(\frac{1}{4}) (x - 3) = 5 - x

2 + 23 i

What are the solutions to the equation (x-2)(x+1)=5 ?
The solutions to the equation (x-2)(x+1)=5 are x=(1+sqrt(29))/2 ,x=(1-sqrt(29))/2
Find the zeros of (x-2)(x+1)=5
The zeros of (x-2)(x+1)=5 are x=(1+sqrt(29))/2 ,x=(1-sqrt(29))/2