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

The solutions to the equation 2x^2-9x+5=0 are x=(9+sqrt(41))/4 ,x=(9-sqrt(41))/4

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

The zeros of 2x^2-9x+5=0 are x=(9+sqrt(41))/4 ,x=(9-sqrt(41))/4

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

x = \frac{9 + 41}{4}, x = \frac{9 - 41}{4}

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- ( - 9 ) + 41}{2 \cdot 2} : \frac{9 + 41}{4}

x = \frac{- ( - 9 ) - 41}{2 \cdot 2} : \frac{9 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{9 + 41}{4}, x = \frac{9 - 41}{4}

\frac{x - 1}{3} = 3

\frac{a ^{3} b ^{5}}{a b ^{2}}

(x - 6) (x - 3) (x + 1) > 0

(3 - 42) (5 - 62)

- 6 + 9

What are the solutions to the equation 2x^2-9x+5=0 ?
The solutions to the equation 2x^2-9x+5=0 are x=(9+sqrt(41))/4 ,x=(9-sqrt(41))/4
Find the zeros of 2x^2-9x+5=0
The zeros of 2x^2-9x+5=0 are x=(9+sqrt(41))/4 ,x=(9-sqrt(41))/4