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

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

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

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

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2 y = 5 y^{2} - 3

y = 1, y = - \frac{3}{5}

Decimal Notation

y = 1, y = - 0.6

Solution steps

2 y = 5 y^{2} - 3

Switch sides

5 y^{2} - 3 = 2 y

Move

2 y

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 5 (- 3) = 8

y_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- ( - 2 ) + 8}{2 \cdot 5}, y_{2} = \frac{- ( - 2 ) - 8}{2 \cdot 5}

y = \frac{- ( - 2 ) + 8}{2 \cdot 5} : 1

y = \frac{- ( - 2 ) - 8}{2 \cdot 5} : - \frac{3}{5}

The solutions to the quadratic equation are:

y = 1, y = - \frac{3}{5}

- 81 x + 19 y - 30 z + 6 y + 80 x + x - 25 y

\frac{12 y ^{5} m}{4 ( y ^{3} m ^{4} ) ^{- 2}}

\frac{x}{5} + 9 = 12

49 x - x^{3} \geq 0

- x^{3} + 5 x^{2} - x + 5

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