What are the solutions to the equation x(x-3)+1=0 ?

The solutions to the equation x(x-3)+1=0 are x=(3+sqrt(5))/2 ,x=(3-sqrt(5))/2

Find the zeros of x(x-3)+1=0

The zeros of x(x-3)+1=0 are x=(3+sqrt(5))/2 ,x=(3-sqrt(5))/2

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x (x - 3) + 1 = 0

x = \frac{3 + 5}{2}, x = \frac{3 - 5}{2}

Decimal Notation

x = 2.61803 \dots, x = 0.38196 \dots

Solution steps

x (x - 3) + 1 = 0

Expand

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

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 = 5

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

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

x = \frac{- ( - 3 ) + 5}{2 \cdot 1} : \frac{3 + 5}{2}

x = \frac{- ( - 3 ) - 5}{2 \cdot 1} : \frac{3 - 5}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 5}{2}, x = \frac{3 - 5}{2}

x, f (x + 3) = 4 x^{2} + 3 x - 4

x, 1 - 4 x + 2 x^{2} - 2 y^{3} = 0

y, y^{2} = 2 y

5.8 \cdot 1 5^{2} + 0.32 \cdot y^{2} - 47 \cdot 15 - 2.7 y + 1750 = 145000

x^{2} + 0.19 x - 0.1197 = 0

What are the solutions to the equation x(x-3)+1=0 ?
The solutions to the equation x(x-3)+1=0 are x=(3+sqrt(5))/2 ,x=(3-sqrt(5))/2
Find the zeros of x(x-3)+1=0
The zeros of x(x-3)+1=0 are x=(3+sqrt(5))/2 ,x=(3-sqrt(5))/2