What are the solutions to the equation x^2-4x+1=0 ?

The solutions to the equation x^2-4x+1=0 are x=2+sqrt(3),x=2-sqrt(3)

Find the zeros of x^2-4x+1=0

The zeros of x^2-4x+1=0 are x=2+sqrt(3),x=2-sqrt(3)

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x^{2} - 4 x + 1 = 0

x = 2 + 3, x = 2 - 3

Decimal Notation

x = 3.73205 \dots, x = 0.26794 \dots

Solution steps

x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 1 = 23

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 3}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 3}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2 3}{2 \cdot 1} : 2 + 3

x = \frac{- ( - 4 ) - 2 3}{2 \cdot 1} : 2 - 3

The solutions to the quadratic equation are:

x = 2 + 3, x = 2 - 3

3 x^{3} + 3 x y - 3 y^{3}

\frac{12 x ^{6} y ^{5}}{8 x ^{4} y ^{2}}

x^{\frac{2}{3}} \cdot x^{2}

∣ 7 x - 5 ∣ + 2 \geq 9

15 h - 2 h - 7 h - 6 h + 2 h + 1 = 13

What are the solutions to the equation x^2-4x+1=0 ?
The solutions to the equation x^2-4x+1=0 are x=2+sqrt(3),x=2-sqrt(3)
Find the zeros of x^2-4x+1=0
The zeros of x^2-4x+1=0 are x=2+sqrt(3),x=2-sqrt(3)