What are the solutions to the equation (17-y)^2+y^2=145 ?

The solutions to the equation (17-y)^2+y^2=145 are y=9,y=8

Find the zeros of (17-y)^2+y^2=145

The zeros of (17-y)^2+y^2=145 are y=9,y=8

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(17 - y)^{2} + y^{2} = 145

y = 9, y = 8

Solution steps

(17 - y)^{2} + y^{2} = 145

Expand

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

289 - 34 y + 2 y^{2} = 145

Move

145

to the left side

2 y^{2} - 34 y + 144 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 34 ) \pm ( - 34 ) ^{2} - 4 \cdot 2 \cdot 144}{2 \cdot 2}

(- 34)^{2} - 4 \cdot 2 \cdot 144 = 2

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

Separate the solutions

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

y = \frac{- ( - 34 ) + 2}{2 \cdot 2} : 9

y = \frac{- ( - 34 ) - 2}{2 \cdot 2} : 8

The solutions to the quadratic equation are:

y = 9, y = 8

5 (3 x + 8)^{2} - 16 = 29

x^{2} - 8 x - 9 = - 9

2 y (y - 3) = 5 (2 y - 1) - 2

x^{2} - 7 x + 10 = 5 x - x^{2}

2 x^{2} - 4 x = 3 x + 4

What are the solutions to the equation (17-y)^2+y^2=145 ?
The solutions to the equation (17-y)^2+y^2=145 are y=9,y=8
Find the zeros of (17-y)^2+y^2=145
The zeros of (17-y)^2+y^2=145 are y=9,y=8