What are the solutions to the equation 5(3-2y)4y=3 ?

The solutions to the equation 5(3-2y)4y=3 are y=(15-sqrt(195))/(20),y=(15+sqrt(195))/(20)

Find the zeros of 5(3-2y)4y=3

The zeros of 5(3-2y)4y=3 are y=(15-sqrt(195))/(20),y=(15+sqrt(195))/(20)

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5 (3 - 2 y) 4 y = 3

y = \frac{15 - 195}{20}, y = \frac{15 + 195}{20}

Decimal Notation

y = 0.05178 \dots, y = 1.44821 \dots

Solution steps

5 (3 - 2 y) \cdot 4 y = 3

Expand

5 (3 - 2 y) \cdot 4 y : 60 y - 40 y^{2}

60 y - 40 y^{2} = 3

Move

3

to the left side

60 y - 40 y^{2} - 3 = 0

Write in the standard form

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

- 40 y^{2} + 60 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 60 \pm 6 0 ^{2} - 4 ( - 40 ) ( - 3 )}{2 ( - 40 )}

6 0^{2} - 4 (- 40) (- 3) = 4195

y_{1, 2} = \frac{- 60 \pm 4 195}{2 ( - 40 )}

Separate the solutions

y_{1} = \frac{- 60 + 4 195}{2 ( - 40 )}, y_{2} = \frac{- 60 - 4 195}{2 ( - 40 )}

y = \frac{- 60 + 4 195}{2 ( - 40 )} : \frac{15 - 195}{20}

y = \frac{- 60 - 4 195}{2 ( - 40 )} : \frac{15 + 195}{20}

The solutions to the quadratic equation are:

y = \frac{15 - 195}{20}, y = \frac{15 + 195}{20}

(x - 5)^{2} + (x - 5) = 0

x, - x^{2} - 3 y^{3} = 5

2 x^{2} = 20 - 3 x^{2}

x, x^{2} + 8 x - 2 = 0

x^{2} + 18 x + 63 = 0

What are the solutions to the equation 5(3-2y)4y=3 ?
The solutions to the equation 5(3-2y)4y=3 are y=(15-sqrt(195))/(20),y=(15+sqrt(195))/(20)
Find the zeros of 5(3-2y)4y=3
The zeros of 5(3-2y)4y=3 are y=(15-sqrt(195))/(20),y=(15+sqrt(195))/(20)