What are the solutions to the equation (3y-10)^2+y^2=10 ?

The solutions to the equation (3y-10)^2+y^2=10 are y=3

Find the zeros of (3y-10)^2+y^2=10

The zeros of (3y-10)^2+y^2=10 are y=3

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

y = 3

Solution steps

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

Expand

(3 y - 10)^{2} + y^{2} : 10 y^{2} - 60 y + 100

10 y^{2} - 60 y + 100 = 10

Move

10

to the left side

10 y^{2} - 60 y + 90 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 60 ) \pm ( - 60 ) ^{2} - 4 \cdot 10 \cdot 90}{2 \cdot 10}

(- 60)^{2} - 4 \cdot 10 \cdot 90 = 0

y_{1, 2} = \frac{- ( - 60 ) \pm 0}{2 \cdot 10}

y = \frac{- ( - 60 )}{2 \cdot 10}

\frac{- ( - 60 )}{2 \cdot 10} = 3

y = 3

The solution to the quadratic equation is:

y = 3

x, - 3 x y^{2} - 2 x^{2} y + 8 x y = 0

- 4.9 x^{2} - 5 x + 450 = 0

7 x^{2} - 3 = 109

5 x^{2} - 8 x = 6

- x^{2} + 15 x + 100 = 0

What are the solutions to the equation (3y-10)^2+y^2=10 ?
The solutions to the equation (3y-10)^2+y^2=10 are y=3
Find the zeros of (3y-10)^2+y^2=10
The zeros of (3y-10)^2+y^2=10 are y=3