What are the solutions to the equation 2((4y^2+41)/5)+3y^2=21 ?

The solutions to the equation 2((4y^2+41)/5)+3y^2=21 are y=1,y=-1

Find the zeros of 2((4y^2+41)/5)+3y^2=21

The zeros of 2((4y^2+41)/5)+3y^2=21 are y=1,y=-1

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2 (\frac{4 y ^{2} + 41}{5}) + 3 y^{2} = 21

y = 1, y = - 1

Solution steps

2 (\frac{4 y ^{2} + 41}{5}) + 3 y^{2} = 21

Expand

2 (\frac{4 y ^{2} + 41}{5}) + 3 y^{2} : \frac{23 y ^{2}}{5} + \frac{82}{5}

\frac{23 y ^{2}}{5} + \frac{82}{5} = 21

Multiply both sides by

5

23 y^{2} + 82 = 105

Move

82

to the right side

23 y^{2} = 23

Divide both sides by

23

y^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

y = 1, y = - 1

1 = 1

- 1 = - 1

y = 1, y = - 1

25 x^{2} + 10 x - 6 = 0

(- 4)^{2}

8 m^{2} - 5 m - 3

7 - ∣ 2 x + 6 ∣ \leq 5

- 3 x^{2} + 10 x - 8

What are the solutions to the equation 2((4y^2+41)/5)+3y^2=21 ?
The solutions to the equation 2((4y^2+41)/5)+3y^2=21 are y=1,y=-1
Find the zeros of 2((4y^2+41)/5)+3y^2=21
The zeros of 2((4y^2+41)/5)+3y^2=21 are y=1,y=-1