What are the solutions to the equation (y-3)^2=4y-12 ?

The solutions to the equation (y-3)^2=4y-12 are y=7,y=3

Find the zeros of (y-3)^2=4y-12

The zeros of (y-3)^2=4y-12 are y=7,y=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(y - 3)^{2} = 4 y - 12

y = 7, y = 3

Solution steps

(y - 3)^{2} = 4 y - 12

Expand

(y - 3)^{2} : y^{2} - 6 y + 9

y^{2} - 6 y + 9 = 4 y - 12

Move

12

to the left side

y^{2} - 6 y + 21 = 4 y

Move

4 y

to the left side

y^{2} - 10 y + 21 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 21}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 21 = 4

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

Separate the solutions

y_{1} = \frac{- ( - 10 ) + 4}{2 \cdot 1}, y_{2} = \frac{- ( - 10 ) - 4}{2 \cdot 1}

y = \frac{- ( - 10 ) + 4}{2 \cdot 1} : 7

y = \frac{- ( - 10 ) - 4}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

y = 7, y = 3

x, x^{2} - 5 x - 6 = 0

7 x^{2} - 29 x = - 4

2 X^{2} - 8 = 0

v^{2} + 8 v + 7 = 0

2 + 24.5 x - 4.9 x^{2} = 0

What are the solutions to the equation (y-3)^2=4y-12 ?
The solutions to the equation (y-3)^2=4y-12 are y=7,y=3
Find the zeros of (y-3)^2=4y-12
The zeros of (y-3)^2=4y-12 are y=7,y=3