What are the solutions to the equation 5y=9y^2-4 ?

The solutions to the equation 5y=9y^2-4 are y=1,y=-4/9

Find the zeros of 5y=9y^2-4

The zeros of 5y=9y^2-4 are y=1,y=-4/9

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5 y = 9 y^{2} - 4

y = 1, y = - \frac{4}{9}

Solution steps

5 y = 9 y^{2} - 4

Switch sides

9 y^{2} - 4 = 5 y

Move

5 y

to the left side

9 y^{2} - 4 - 5 y = 0

Write in the standard form

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

9 y^{2} - 5 y - 4 = 0

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 9 (- 4) = 13

y_{1, 2} = \frac{- ( - 5 ) \pm 13}{2 \cdot 9}

Separate the solutions

y_{1} = \frac{- ( - 5 ) + 13}{2 \cdot 9}, y_{2} = \frac{- ( - 5 ) - 13}{2 \cdot 9}

y = \frac{- ( - 5 ) + 13}{2 \cdot 9} : 1

y = \frac{- ( - 5 ) - 13}{2 \cdot 9} : - \frac{4}{9}

The solutions to the quadratic equation are:

y = 1, y = - \frac{4}{9}

(1 - i) (1 + i)

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2 x^{4} - 4 x^{2} - 16 \leq 0

6188 n^{2} + 1105 n - 17017

- x^{2} + 3 x = 0

What are the solutions to the equation 5y=9y^2-4 ?
The solutions to the equation 5y=9y^2-4 are y=1,y=-4/9
Find the zeros of 5y=9y^2-4
The zeros of 5y=9y^2-4 are y=1,y=-4/9