What are the solutions to the equation (5-3y)^2+y^2=25 ?

The solutions to the equation (5-3y)^2+y^2=25 are y=3,y=0

Find the zeros of (5-3y)^2+y^2=25

The zeros of (5-3y)^2+y^2=25 are y=3,y=0

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(5 - 3 y)^{2} + y^{2} = 25

y = 3, y = 0

Solution steps

(5 - 3 y)^{2} + y^{2} = 25

Expand

(5 - 3 y)^{2} + y^{2} : 25 - 30 y + 10 y^{2}

25 - 30 y + 10 y^{2} = 25

Move

25

to the left side

10 y^{2} - 30 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 \cdot 10 \cdot 0}{2 \cdot 10}

(- 30)^{2} - 4 \cdot 10 \cdot 0 = 30

y_{1, 2} = \frac{- ( - 30 ) \pm 30}{2 \cdot 10}

Separate the solutions

y_{1} = \frac{- ( - 30 ) + 30}{2 \cdot 10}, y_{2} = \frac{- ( - 30 ) - 30}{2 \cdot 10}

y = \frac{- ( - 30 ) + 30}{2 \cdot 10} : 3

y = \frac{- ( - 30 ) - 30}{2 \cdot 10} : 0

The solutions to the quadratic equation are:

y = 3, y = 0

3 n^{2} - n - 6 = 0

s^{2} - s = 0

16.7 - \frac{20 x ^{2}}{10} = 0

- x^{2} + 10 x + 25 = 0

10 = 3 Q^{2} - 18 Q + 5

What are the solutions to the equation (5-3y)^2+y^2=25 ?
The solutions to the equation (5-3y)^2+y^2=25 are y=3,y=0
Find the zeros of (5-3y)^2+y^2=25
The zeros of (5-3y)^2+y^2=25 are y=3,y=0