What are the solutions to the equation (-3y)^2+y^2=40 ?

The solutions to the equation (-3y)^2+y^2=40 are y=2,y=-2

Find the zeros of (-3y)^2+y^2=40

The zeros of (-3y)^2+y^2=40 are y=2,y=-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(- 3 y)^{2} + y^{2} = 40

y = 2, y = - 2

Solution steps

(- 3 y)^{2} + y^{2} = 40

Expand

(- 3 y)^{2} + y^{2} : 10 y^{2}

10 y^{2} = 40

Move

40

to the left side

10 y^{2} - 40 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 10 ( - 40 )}{2 \cdot 10}

0^{2} - 4 \cdot 10 (- 40) = 40

y_{1, 2} = \frac{- 0 \pm 40}{2 \cdot 10}

Separate the solutions

y_{1} = \frac{- 0 + 40}{2 \cdot 10}, y_{2} = \frac{- 0 - 40}{2 \cdot 10}

y = \frac{- 0 + 40}{2 \cdot 10} : 2

y = \frac{- 0 - 40}{2 \cdot 10} : - 2

The solutions to the quadratic equation are:

y = 2, y = - 2

4 x + x \cdot 4 (x - 2) = (3 x + 15) (x - 2)

16 t^{2} - 64 t + 48 = 0

2 x^{2} - 3 x = - 6

a, u^{3} = a^{2} t

x^{2} + 20 x = 400

What are the solutions to the equation (-3y)^2+y^2=40 ?
The solutions to the equation (-3y)^2+y^2=40 are y=2,y=-2
Find the zeros of (-3y)^2+y^2=40
The zeros of (-3y)^2+y^2=40 are y=2,y=-2