What are the solutions to the equation (21-y)^2+y^2=221 ?

The solutions to the equation (21-y)^2+y^2=221 are y=11,y=10

Find the zeros of (21-y)^2+y^2=221

The zeros of (21-y)^2+y^2=221 are y=11,y=10

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(21 - y)^{2} + y^{2} = 221

y = 11, y = 10

Solution steps

(21 - y)^{2} + y^{2} = 221

Expand

(21 - y)^{2} + y^{2} : 441 - 42 y + 2 y^{2}

441 - 42 y + 2 y^{2} = 221

Move

221

to the left side

2 y^{2} - 42 y + 220 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 42 ) \pm ( - 42 ) ^{2} - 4 \cdot 2 \cdot 220}{2 \cdot 2}

(- 42)^{2} - 4 \cdot 2 \cdot 220 = 2

y_{1, 2} = \frac{- ( - 42 ) \pm 2}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 42 ) + 2}{2 \cdot 2}, y_{2} = \frac{- ( - 42 ) - 2}{2 \cdot 2}

y = \frac{- ( - 42 ) + 2}{2 \cdot 2} : 11

y = \frac{- ( - 42 ) - 2}{2 \cdot 2} : 10

The solutions to the quadratic equation are:

y = 11, y = 10

4.6 x = 4 - x^{2}

- 12 x^{2} + 12 = 0

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

s, h (s) = - 16 s^{2} + 80 s

a = x^{2} - 18 x + 80

What are the solutions to the equation (21-y)^2+y^2=221 ?
The solutions to the equation (21-y)^2+y^2=221 are y=11,y=10
Find the zeros of (21-y)^2+y^2=221
The zeros of (21-y)^2+y^2=221 are y=11,y=10