What are the solutions to the equation (y+10-3)^2+y^2=49 ?

The solutions to the equation (y+10-3)^2+y^2=49 are y=0,y=-7

Find the zeros of (y+10-3)^2+y^2=49

The zeros of (y+10-3)^2+y^2=49 are y=0,y=-7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(y + 10 - 3)^{2} + y^{2} = 49

y = 0, y = - 7

Solution steps

(y + 10 - 3)^{2} + y^{2} = 49

Expand

(y + 10 - 3)^{2} + y^{2} : 2 y^{2} + 14 y + 49

2 y^{2} + 14 y + 49 = 49

Move

49

to the left side

2 y^{2} + 14 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

1 4^{2} - 4 \cdot 2 \cdot 0 = 14

y_{1, 2} = \frac{- 14 \pm 14}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 14 + 14}{2 \cdot 2}, y_{2} = \frac{- 14 - 14}{2 \cdot 2}

y = \frac{- 14 + 14}{2 \cdot 2} : 0

y = \frac{- 14 - 14}{2 \cdot 2} : - 7

The solutions to the quadratic equation are:

y = 0, y = - 7

\frac{4 - ( \frac{3}{4} ) ^{2}}{2 ( \frac{3}{4} )}

\frac{4 n}{4 n - 4} \cdot \frac{n - 1}{n + 1}

4 \cdot (\frac{1}{2})^{1}

z^{8} z^{5}

What are the solutions to the equation (y+10-3)^2+y^2=49 ?
The solutions to the equation (y+10-3)^2+y^2=49 are y=0,y=-7
Find the zeros of (y+10-3)^2+y^2=49
The zeros of (y+10-3)^2+y^2=49 are y=0,y=-7