What are the solutions to the equation 0=x^2-3x-1 ?

The solutions to the equation 0=x^2-3x-1 are x=(3+sqrt(13))/2 ,x=(3-sqrt(13))/2

Find the zeros of 0=x^2-3x-1

The zeros of 0=x^2-3x-1 are x=(3+sqrt(13))/2 ,x=(3-sqrt(13))/2

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0 = 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

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

Switch sides

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 - x^{2} + 42

16 (3920 + 6.25 x) (1000 - x) = 125

- \frac{3}{2} x \geq - 6

(625 x^{6})^{\frac{3}{4}}

4 x + 2 x + 6 = 90

What are the solutions to the equation 0=x^2-3x-1 ?
The solutions to the equation 0=x^2-3x-1 are x=(3+sqrt(13))/2 ,x=(3-sqrt(13))/2
Find the zeros of 0=x^2-3x-1
The zeros of 0=x^2-3x-1 are x=(3+sqrt(13))/2 ,x=(3-sqrt(13))/2