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 = - 0.30277 \dots, x = 3.30277 \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 ( - 1 ) \cdot 1}{2 ( - 1 )}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

\frac{1}{12.5} + \frac{1}{45.5} + \frac{1}{85.5}

\frac{k ^{\frac{1}{3}}}{k ^{\frac{2}{3}} \cdot k ^{- 1}}

6 5^{2}

\frac{y ^{3}}{y ^{2}}

16 x^{2} + 62 x + 42 = 0

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