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

The solutions to the equation 5x^2-9x+2=0 are x=(9+sqrt(41))/(10),x=(9-sqrt(41))/(10)

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

The zeros of 5x^2-9x+2=0 are x=(9+sqrt(41))/(10),x=(9-sqrt(41))/(10)

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

x = \frac{9 + 41}{10}, x = \frac{9 - 41}{10}

Solution steps

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

Solve with the quadratic formula

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

(- 9)^{2} - 4 \cdot 5 \cdot 2 = 41

x_{1, 2} = \frac{- ( - 9 ) \pm 41}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 41}{2 \cdot 5}, x_{2} = \frac{- ( - 9 ) - 41}{2 \cdot 5}

x = \frac{- ( - 9 ) + 41}{2 \cdot 5} : \frac{9 + 41}{10}

x = \frac{- ( - 9 ) - 41}{2 \cdot 5} : \frac{9 - 41}{10}

The solutions to the quadratic equation are:

x = \frac{9 + 41}{10}, x = \frac{9 - 41}{10}

18 x y^{2} + 12 x y + 2 x

\frac{∣ 2 - x ∣}{∣ x + 8 ∣} \geq 0

- 0.5244 = \frac{x - 45}{14}

\frac{2 x}{5} - x = \frac{x}{15} - \frac{4}{3}

\frac{π}{2} - \frac{π}{3}

What are the solutions to the equation 5x^2-9x+2=0 ?
The solutions to the equation 5x^2-9x+2=0 are x=(9+sqrt(41))/(10),x=(9-sqrt(41))/(10)
Find the zeros of 5x^2-9x+2=0
The zeros of 5x^2-9x+2=0 are x=(9+sqrt(41))/(10),x=(9-sqrt(41))/(10)