What are the solutions to the equation (y-2)^2=-y^2+10 ?

The solutions to the equation (y-2)^2=-y^2+10 are y=3,y=-1

Find the zeros of (y-2)^2=-y^2+10

The zeros of (y-2)^2=-y^2+10 are y=3,y=-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(y - 2)^{2} = - y^{2} + 10

y = 3, y = - 1

Solution steps

(y - 2)^{2} = - y^{2} + 10

Expand

(y - 2)^{2} : y^{2} - 4 y + 4

y^{2} - 4 y + 4 = - y^{2} + 10

Move

10

to the left side

y^{2} - 4 y - 6 = - y^{2}

Move

y^{2}

to the left side

2 y^{2} - 4 y - 6 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 (- 6) = 8

y_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 2}, y_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 2}

y = \frac{- ( - 4 ) + 8}{2 \cdot 2} : 3

y = \frac{- ( - 4 ) - 8}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

y = 3, y = - 1

r^{2} + 5 r - 3 = 0

6 x^{2} = 384

x, x^{2} - 6 x + 9 - 4 y^{2} = 0

0 = 3 k^{2} + 6000 k - 0.25 \cdot 1 0^{6}

x, x^{2} + y^{2} = cy

What are the solutions to the equation (y-2)^2=-y^2+10 ?
The solutions to the equation (y-2)^2=-y^2+10 are y=3,y=-1
Find the zeros of (y-2)^2=-y^2+10
The zeros of (y-2)^2=-y^2+10 are y=3,y=-1