What are the solutions to the equation (9-y)^2+y^2=53 ?

The solutions to the equation (9-y)^2+y^2=53 are y=7,y=2

Find the zeros of (9-y)^2+y^2=53

The zeros of (9-y)^2+y^2=53 are y=7,y=2

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(9 - y)^{2} + y^{2} = 53

y = 7, y = 2

Solution steps

(9 - y)^{2} + y^{2} = 53

Expand

(9 - y)^{2} + y^{2} : 81 - 18 y + 2 y^{2}

81 - 18 y + 2 y^{2} = 53

Move

53

to the left side

2 y^{2} - 18 y + 28 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 2 \cdot 28}{2 \cdot 2}

(- 18)^{2} - 4 \cdot 2 \cdot 28 = 10

y_{1, 2} = \frac{- ( - 18 ) \pm 10}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 18 ) + 10}{2 \cdot 2}, y_{2} = \frac{- ( - 18 ) - 10}{2 \cdot 2}

y = \frac{- ( - 18 ) + 10}{2 \cdot 2} : 7

y = \frac{- ( - 18 ) - 10}{2 \cdot 2} : 2

The solutions to the quadratic equation are:

y = 7, y = 2

m^{2} - 1.04 = 1.16

8 x^{2} + x - 9 = 0

- 2 x^{2} + 3 x + 4 = 0

48 = (2 x + 6) (2 x + 4)

x^{2} - 16 x = 21

What are the solutions to the equation (9-y)^2+y^2=53 ?
The solutions to the equation (9-y)^2+y^2=53 are y=7,y=2
Find the zeros of (9-y)^2+y^2=53
The zeros of (9-y)^2+y^2=53 are y=7,y=2