What are the solutions to the equation 3y^2-5y+2=0 ?

The solutions to the equation 3y^2-5y+2=0 are y=1,y= 2/3

Find the zeros of 3y^2-5y+2=0

The zeros of 3y^2-5y+2=0 are y=1,y= 2/3

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3 y^{2} - 5 y + 2 = 0

y = 1, y = \frac{2}{3}

Solution steps

3 y^{2} - 5 y + 2 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 \cdot 2}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 \cdot 2 = 1

y_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 3}, y_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 3}

y = \frac{- ( - 5 ) + 1}{2 \cdot 3} : 1

y = \frac{- ( - 5 ) - 1}{2 \cdot 3} : \frac{2}{3}

The solutions to the quadratic equation are:

y = 1, y = \frac{2}{3}

r^{2} - r - 1 = 0

x^{2} - x = 9

3 x^{2} + x + 7 = x^{2} + 10

4 x^{2} = - 1

7 - 7 x = (3 x + 2) (x - 1)

What are the solutions to the equation 3y^2-5y+2=0 ?
The solutions to the equation 3y^2-5y+2=0 are y=1,y= 2/3
Find the zeros of 3y^2-5y+2=0
The zeros of 3y^2-5y+2=0 are y=1,y= 2/3