What are the solutions to the equation 2x^2-5x+1=0 ?

The solutions to the equation 2x^2-5x+1=0 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4

Find the zeros of 2x^2-5x+1=0

The zeros of 2x^2-5x+1=0 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4

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2 x^{2} - 5 x + 1 = 0

x = \frac{5 + 17}{4}, x = \frac{5 - 17}{4}

Decimal Notation

x = 2.28077 \dots, x = 0.21922 \dots

Solution steps

2 x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 2 \cdot 1 = 17

x_{1, 2} = \frac{- ( - 5 ) \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 17}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 17}{2 \cdot 2}

x = \frac{- ( - 5 ) + 17}{2 \cdot 2} : \frac{5 + 17}{4}

x = \frac{- ( - 5 ) - 17}{2 \cdot 2} : \frac{5 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 17}{4}, x = \frac{5 - 17}{4}

\frac{- 2}{5 ^{2}}

8 - 5 y = - 3 y - 31

\frac{380}{25} \cdot 151

- 7 = \frac{y}{7}

ln (6 a + 9 b)

What are the solutions to the equation 2x^2-5x+1=0 ?
The solutions to the equation 2x^2-5x+1=0 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4
Find the zeros of 2x^2-5x+1=0
The zeros of 2x^2-5x+1=0 are x=(5+sqrt(17))/4 ,x=(5-sqrt(17))/4