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

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

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

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

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3 y^{2} + 2 = 5 y

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

Solution steps

3 y^{2} + 2 = 5 y

Move

5 y

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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}

\frac{( 2 x + 3 ) ^{2}}{2} = - 1

9 x^{2} - 8 = 0

x, x^{2} y^{2} + x y = 1

25 - r^{2} = 4 r

x^{2} + 4 = 2 x^{2} - 12

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