What are the solutions to the equation 4x^2-6x+1=0 ?

The solutions to the equation 4x^2-6x+1=0 are x=(3+sqrt(5))/4 ,x=(3-sqrt(5))/4

Find the zeros of 4x^2-6x+1=0

The zeros of 4x^2-6x+1=0 are x=(3+sqrt(5))/4 ,x=(3-sqrt(5))/4

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4 x^{2} - 6 x + 1 = 0

x = \frac{3 + 5}{4}, x = \frac{3 - 5}{4}

Decimal Notation

x = 1.30901 \dots, x = 0.19098 \dots

Solution steps

4 x^{2} - 6 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

(- 6)^{2} - 4 \cdot 4 \cdot 1 = 25

x_{1, 2} = \frac{- ( - 6 ) \pm 2 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 5}{2 \cdot 4}, x_{2} = \frac{- ( - 6 ) - 2 5}{2 \cdot 4}

x = \frac{- ( - 6 ) + 2 5}{2 \cdot 4} : \frac{3 + 5}{4}

x = \frac{- ( - 6 ) - 2 5}{2 \cdot 4} : \frac{3 - 5}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 5}{4}, x = \frac{3 - 5}{4}

8 0^{\frac{1}{4}}

12 x^{2} + 50 x y + 28 y^{2}

x - 4 > 2 x + 3

24 < 4 + 2 (h - 2) or 4 + 2 (h - 2) < 32

x^{2} - x - 12 < x

What are the solutions to the equation 4x^2-6x+1=0 ?
The solutions to the equation 4x^2-6x+1=0 are x=(3+sqrt(5))/4 ,x=(3-sqrt(5))/4
Find the zeros of 4x^2-6x+1=0
The zeros of 4x^2-6x+1=0 are x=(3+sqrt(5))/4 ,x=(3-sqrt(5))/4