What are the solutions to the equation (18-y)^2+y^2=180 ?

The solutions to the equation (18-y)^2+y^2=180 are y=12,y=6

Find the zeros of (18-y)^2+y^2=180

The zeros of (18-y)^2+y^2=180 are y=12,y=6

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(18 - y)^{2} + y^{2} = 180

y = 12, y = 6

Solution steps

(18 - y)^{2} + y^{2} = 180

Expand

(18 - y)^{2} + y^{2} : 324 - 36 y + 2 y^{2}

324 - 36 y + 2 y^{2} = 180

Move

180

to the left side

2 y^{2} - 36 y + 144 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 2 \cdot 144}{2 \cdot 2}

(- 36)^{2} - 4 \cdot 2 \cdot 144 = 12

y_{1, 2} = \frac{- ( - 36 ) \pm 12}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 36 ) + 12}{2 \cdot 2}, y_{2} = \frac{- ( - 36 ) - 12}{2 \cdot 2}

y = \frac{- ( - 36 ) + 12}{2 \cdot 2} : 12

y = \frac{- ( - 36 ) - 12}{2 \cdot 2} : 6

The solutions to the quadratic equation are:

y = 12, y = 6

5 x^{2} - 9 = 0

y, (16 - x) y^{2} = x^{3}

w^{2} - 8 w + 15 = 0

t, f (6 - t) = 3 - 5 (6 - t) - 2 (6 - t)^{2}

t, s = 32 t - 16 t^{2}

What are the solutions to the equation (18-y)^2+y^2=180 ?
The solutions to the equation (18-y)^2+y^2=180 are y=12,y=6
Find the zeros of (18-y)^2+y^2=180
The zeros of (18-y)^2+y^2=180 are y=12,y=6