What are the solutions to the equation (y-3)^2=3y-5 ?

The solutions to the equation (y-3)^2=3y-5 are y=7,y=2

Find the zeros of (y-3)^2=3y-5

The zeros of (y-3)^2=3y-5 are y=7,y=2

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(y - 3)^{2} = 3 y - 5

y = 7, y = 2

Solution steps

(y - 3)^{2} = 3 y - 5

Expand

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

y^{2} - 6 y + 9 = 3 y - 5

Move

5

to the left side

y^{2} - 6 y + 14 = 3 y

Move

3 y

to the left side

y^{2} - 9 y + 14 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 14}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 14 = 5

y_{1, 2} = \frac{- ( - 9 ) \pm 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 9 ) + 5}{2 \cdot 1}, y_{2} = \frac{- ( - 9 ) - 5}{2 \cdot 1}

y = \frac{- ( - 9 ) + 5}{2 \cdot 1} : 7

y = \frac{- ( - 9 ) - 5}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

y = 7, y = 2

x^{2} - 8 x + 16 = 4

\frac{4}{3} π (7)^{3}

\frac{2 ( 5 )}{( 5 + 1 ) ^{2}}

3 a^{2} + 8 ab + 5 b^{2}

3 x + 2 < 20

What are the solutions to the equation (y-3)^2=3y-5 ?
The solutions to the equation (y-3)^2=3y-5 are y=7,y=2
Find the zeros of (y-3)^2=3y-5
The zeros of (y-3)^2=3y-5 are y=7,y=2