What are the solutions to the equation (15-y)^2+y^2=125 ?

The solutions to the equation (15-y)^2+y^2=125 are y=10,y=5

Find the zeros of (15-y)^2+y^2=125

The zeros of (15-y)^2+y^2=125 are y=10,y=5

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(15 - y)^{2} + y^{2} = 125

y = 10, y = 5

Solution steps

(15 - y)^{2} + y^{2} = 125

Expand

(15 - y)^{2} + y^{2} : 225 - 30 y + 2 y^{2}

225 - 30 y + 2 y^{2} = 125

Move

125

to the left side

2 y^{2} - 30 y + 100 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 \cdot 2 \cdot 100}{2 \cdot 2}

(- 30)^{2} - 4 \cdot 2 \cdot 100 = 10

y_{1, 2} = \frac{- ( - 30 ) \pm 10}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 30 ) + 10}{2 \cdot 2}, y_{2} = \frac{- ( - 30 ) - 10}{2 \cdot 2}

y = \frac{- ( - 30 ) + 10}{2 \cdot 2} : 10

y = \frac{- ( - 30 ) - 10}{2 \cdot 2} : 5

The solutions to the quadratic equation are:

y = 10, y = 5

x^{2} + 1 = 1

t (8 - 0.8 t) = 0

5 x > 30

10 \geq \frac{x}{3} + 6

2 z \leq - 6

What are the solutions to the equation (15-y)^2+y^2=125 ?
The solutions to the equation (15-y)^2+y^2=125 are y=10,y=5
Find the zeros of (15-y)^2+y^2=125
The zeros of (15-y)^2+y^2=125 are y=10,y=5