What are the solutions to the equation (17-y)^2+y^2=169 ?

The solutions to the equation (17-y)^2+y^2=169 are y=12,y=5

Find the zeros of (17-y)^2+y^2=169

The zeros of (17-y)^2+y^2=169 are y=12,y=5

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(17 - y)^{2} + y^{2} = 169

y = 12, y = 5

Solution steps

(17 - y)^{2} + y^{2} = 169

Expand

(17 - y)^{2} + y^{2} : 289 - 34 y + 2 y^{2}

289 - 34 y + 2 y^{2} = 169

Move

169

to the left side

2 y^{2} - 34 y + 120 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 34 ) \pm ( - 34 ) ^{2} - 4 \cdot 2 \cdot 120}{2 \cdot 2}

(- 34)^{2} - 4 \cdot 2 \cdot 120 = 14

y_{1, 2} = \frac{- ( - 34 ) \pm 14}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 34 ) + 14}{2 \cdot 2}, y_{2} = \frac{- ( - 34 ) - 14}{2 \cdot 2}

y = \frac{- ( - 34 ) + 14}{2 \cdot 2} : 12

y = \frac{- ( - 34 ) - 14}{2 \cdot 2} : 5

The solutions to the quadratic equation are:

y = 12, y = 5

x^{2} + 19 x + 78 = 0

\frac{4}{3} x^{2} = 48

(x + 1)^{2} - 72 = 0

(x - 10)^{2} = 16

x, x^{2} + 8 x = 0

What are the solutions to the equation (17-y)^2+y^2=169 ?
The solutions to the equation (17-y)^2+y^2=169 are y=12,y=5
Find the zeros of (17-y)^2+y^2=169
The zeros of (17-y)^2+y^2=169 are y=12,y=5