What are the solutions to the equation x^2-3x=1 ?

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

Find the zeros of x^2-3x=1

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

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x^{2} - 3 x = 1

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

Decimal Notation

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

Solution steps

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}

\frac{1}{- 1 + y} (- x + \frac{x ^{2}}{2})

\frac{( 3 - 5 i )}{( 2 - 5 i )}

\frac{1}{3} (6 + x) - 5 = 12

4 (x + h)^{2} - 2 (x + h) - (4 x^{2} - 2 x)

- x^{2} - 8 x - 12

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