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

The solutions to the equation x(x-3)=1 are x=(3+sqrt(13))/2 ,x=(3-sqrt(13))/2

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

The zeros of x(x-3)=1 are x=(3+sqrt(13))/2 ,x=(3-sqrt(13))/2

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x (x - 3) = 1

x = \frac{3 + 13}{2}, x = \frac{3 - 13}{2}

Decimal Notation

x = 3.30277 \dots, x = - 0.30277 \dots

Solution steps

x (x - 3) = 1

Expand

x (x - 3) : x^{2} - 3 x

x^{2} - 3 x = 1

Move

1

to the left side

x^{2} - 3 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 1) = 13

x_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 13}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 13}{2 \cdot 1}

x = \frac{- ( - 3 ) + 13}{2 \cdot 1} : \frac{3 + 13}{2}

x = \frac{- ( - 3 ) - 13}{2 \cdot 1} : \frac{3 - 13}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 13}{2}, x = \frac{3 - 13}{2}

x^{2} + 7 x + 1 = - 9

x, x^{2} - 6 x + y^{2} + 8 y + 25 = 0

t, h = 100 t - 16 t^{2}

y = 3 - y^{2}

- 5 b^{2} + 2 b = 0

What are the solutions to the equation x(x-3)=1 ?
The solutions to the equation x(x-3)=1 are x=(3+sqrt(13))/2 ,x=(3-sqrt(13))/2
Find the zeros of x(x-3)=1
The zeros of x(x-3)=1 are x=(3+sqrt(13))/2 ,x=(3-sqrt(13))/2