What are the solutions to the equation x^2-3x+1=6 ?

The solutions to the equation x^2-3x+1=6 are x=(3+sqrt(29))/2 ,x=(3-sqrt(29))/2

Find the zeros of x^2-3x+1=6

The zeros of x^2-3x+1=6 are x=(3+sqrt(29))/2 ,x=(3-sqrt(29))/2

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x^{2} - 3 x + 1 = 6

x = \frac{3 + 29}{2}, x = \frac{3 - 29}{2}

Decimal Notation

x = 4.19258 \dots, x = - 1.19258 \dots

Solution steps

x^{2} - 3 x + 1 = 6

Move

6

to the left side

x^{2} - 3 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 5) = 29

x_{1, 2} = \frac{- ( - 3 ) \pm 29}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 29}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 29}{2 \cdot 1}

x = \frac{- ( - 3 ) + 29}{2 \cdot 1} : \frac{3 + 29}{2}

x = \frac{- ( - 3 ) - 29}{2 \cdot 1} : \frac{3 - 29}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 29}{2}, x = \frac{3 - 29}{2}

3 x^{2} - 8 x + 6 = 17

1.6 \cdot 1 0^{- 19} \cdot 100 = \frac{1}{2} \cdot 1.67 \cdot 1 0^{- 27} v^{2}

- 2 z^{2} + 8 z - 8 = 0

- 16 t^{2} + 65.62 t - 98.43 = 0

(x - 3)^{2} + (x - 1)^{2} = 1

What are the solutions to the equation x^2-3x+1=6 ?
The solutions to the equation x^2-3x+1=6 are x=(3+sqrt(29))/2 ,x=(3-sqrt(29))/2
Find the zeros of x^2-3x+1=6
The zeros of x^2-3x+1=6 are x=(3+sqrt(29))/2 ,x=(3-sqrt(29))/2