What are the solutions to the equation 2x^2-9x+1=0 ?

The solutions to the equation 2x^2-9x+1=0 are x=(9+sqrt(73))/4 ,x=(9-sqrt(73))/4

Find the zeros of 2x^2-9x+1=0

The zeros of 2x^2-9x+1=0 are x=(9+sqrt(73))/4 ,x=(9-sqrt(73))/4

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2 x^{2} - 9 x + 1 = 0

x = \frac{9 + 73}{4}, x = \frac{9 - 73}{4}

Decimal Notation

x = 4.38600 \dots, x = 0.11399 \dots

Solution steps

2 x^{2} - 9 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

(- 9)^{2} - 4 \cdot 2 \cdot 1 = 73

x_{1, 2} = \frac{- ( - 9 ) \pm 73}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 73}{2 \cdot 2}, x_{2} = \frac{- ( - 9 ) - 73}{2 \cdot 2}

x = \frac{- ( - 9 ) + 73}{2 \cdot 2} : \frac{9 + 73}{4}

x = \frac{- ( - 9 ) - 73}{2 \cdot 2} : \frac{9 - 73}{4}

The solutions to the quadratic equation are:

x = \frac{9 + 73}{4}, x = \frac{9 - 73}{4}

5 x^{2} + 25 x + 30

\frac{7}{18} - \frac{7}{2}

- 10 - 10

(\frac{1}{64})^{\frac{1}{6}}

- 3 n + 14 \leq 3 n

What are the solutions to the equation 2x^2-9x+1=0 ?
The solutions to the equation 2x^2-9x+1=0 are x=(9+sqrt(73))/4 ,x=(9-sqrt(73))/4
Find the zeros of 2x^2-9x+1=0
The zeros of 2x^2-9x+1=0 are x=(9+sqrt(73))/4 ,x=(9-sqrt(73))/4