What are the solutions to the equation (x-2)^2-4=x+1 ?

The solutions to the equation (x-2)^2-4=x+1 are x=(5+sqrt(29))/2 ,x=(5-sqrt(29))/2

Find the zeros of (x-2)^2-4=x+1

The zeros of (x-2)^2-4=x+1 are x=(5+sqrt(29))/2 ,x=(5-sqrt(29))/2

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(x - 2)^{2} - 4 = x + 1

x = \frac{5 + 29}{2}, x = \frac{5 - 29}{2}

Decimal Notation

x = 5.19258 \dots, x = - 0.19258 \dots

Solution steps

(x - 2)^{2} - 4 = x + 1

Expand

(x - 2)^{2} - 4 : x^{2} - 4 x

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

Move

1

to the left side

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

Move

x

to the left side

x^{2} - 5 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 1) = 29

x_{1, 2} = \frac{- ( - 5 ) \pm 29}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 29}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 29}{2 \cdot 1}

x = \frac{- ( - 5 ) + 29}{2 \cdot 1} : \frac{5 + 29}{2}

x = \frac{- ( - 5 ) - 29}{2 \cdot 1} : \frac{5 - 29}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 29}{2}, x = \frac{5 - 29}{2}

- \frac{2}{3} \cdot \frac{3}{2}

(x - 3)^{2} = 144

3^{\frac{1}{4} - 1}

\frac{x}{x}

6 h - 5 (h - 1) \leq 7 h - 11

What are the solutions to the equation (x-2)^2-4=x+1 ?
The solutions to the equation (x-2)^2-4=x+1 are x=(5+sqrt(29))/2 ,x=(5-sqrt(29))/2
Find the zeros of (x-2)^2-4=x+1
The zeros of (x-2)^2-4=x+1 are x=(5+sqrt(29))/2 ,x=(5-sqrt(29))/2