What are the solutions to the equation 5x^2-9x+1=0 ?

The solutions to the equation 5x^2-9x+1=0 are x=(9+sqrt(61))/(10),x=(9-sqrt(61))/(10)

Find the zeros of 5x^2-9x+1=0

The zeros of 5x^2-9x+1=0 are x=(9+sqrt(61))/(10),x=(9-sqrt(61))/(10)

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

x = \frac{9 + 61}{10}, x = \frac{9 - 61}{10}

Decimal Notation

x = 1.68102 \dots, x = 0.11897 \dots

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 5 \cdot 1}{2 \cdot 5}

(- 9)^{2} - 4 \cdot 5 \cdot 1 = 61

x_{1, 2} = \frac{- ( - 9 ) \pm 61}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 61}{2 \cdot 5}, x_{2} = \frac{- ( - 9 ) - 61}{2 \cdot 5}

x = \frac{- ( - 9 ) + 61}{2 \cdot 5} : \frac{9 + 61}{10}

x = \frac{- ( - 9 ) - 61}{2 \cdot 5} : \frac{9 - 61}{10}

The solutions to the quadratic equation are:

x = \frac{9 + 61}{10}, x = \frac{9 - 61}{10}

(x)^{4}

\frac{3 ( \frac{π}{2} ) ^{2}}{2}

\frac{x - 3}{4 x + 7} \leq 5

\frac{x}{8} = - 9

3^{- 4} = \frac{1}{81}

What are the solutions to the equation 5x^2-9x+1=0 ?
The solutions to the equation 5x^2-9x+1=0 are x=(9+sqrt(61))/(10),x=(9-sqrt(61))/(10)
Find the zeros of 5x^2-9x+1=0
The zeros of 5x^2-9x+1=0 are x=(9+sqrt(61))/(10),x=(9-sqrt(61))/(10)