What are the solutions to the equation (10-y)^2+y^2=68 ?

The solutions to the equation (10-y)^2+y^2=68 are y=8,y=2

Find the zeros of (10-y)^2+y^2=68

The zeros of (10-y)^2+y^2=68 are y=8,y=2

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(10 - y)^{2} + y^{2} = 68

y = 8, y = 2

Solution steps

(10 - y)^{2} + y^{2} = 68

Expand

(10 - y)^{2} + y^{2} : 100 - 20 y + 2 y^{2}

100 - 20 y + 2 y^{2} = 68

Move

68

to the left side

2 y^{2} - 20 y + 32 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 2 \cdot 32}{2 \cdot 2}

(- 20)^{2} - 4 \cdot 2 \cdot 32 = 12

y_{1, 2} = \frac{- ( - 20 ) \pm 12}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 20 ) + 12}{2 \cdot 2}, y_{2} = \frac{- ( - 20 ) - 12}{2 \cdot 2}

y = \frac{- ( - 20 ) + 12}{2 \cdot 2} : 8

y = \frac{- ( - 20 ) - 12}{2 \cdot 2} : 2

The solutions to the quadratic equation are:

y = 8, y = 2

x^{2} - 6 x + 9 = 2

2 s^{2} + s + 1 = 0

x^{2} + 8 + 3

0.81307831 x^{2} - 71.248 + 1600 = 0

- 4 x^{2} - 28 x = 0

What are the solutions to the equation (10-y)^2+y^2=68 ?
The solutions to the equation (10-y)^2+y^2=68 are y=8,y=2
Find the zeros of (10-y)^2+y^2=68
The zeros of (10-y)^2+y^2=68 are y=8,y=2