What are the solutions to the equation 0^2+y^2+4(0)-10y+4=0 ?

The solutions to the equation 0^2+y^2+4(0)-10y+4=0 are y=5+sqrt(21),y=5-sqrt(21)

Find the zeros of 0^2+y^2+4(0)-10y+4=0

The zeros of 0^2+y^2+4(0)-10y+4=0 are y=5+sqrt(21),y=5-sqrt(21)

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0^{2} + y^{2} + 4 (0) - 10 y + 4 = 0

y = 5 + 21, y = 5 - 21

Decimal Notation

y = 9.58257 \dots, y = 0.41742 \dots

Solution steps

0^{2} + y^{2} + 4 (0) - 10 y + 4 = 0

Expand

0^{2} + y^{2} + 4 (0) - 10 y + 4 : y^{2} - 10 y + 4

y^{2} - 10 y + 4 = 0

Solve with the quadratic formula

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

(- 10)^{2} - 4 \cdot 1 \cdot 4 = 221

y_{1, 2} = \frac{- ( - 10 ) \pm 2 21}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 10 ) + 2 21}{2 \cdot 1}, y_{2} = \frac{- ( - 10 ) - 2 21}{2 \cdot 1}

y = \frac{- ( - 10 ) + 2 21}{2 \cdot 1} : 5 + 21

y = \frac{- ( - 10 ) - 2 21}{2 \cdot 1} : 5 - 21

The solutions to the quadratic equation are:

y = 5 + 21, y = 5 - 21

- x^{2} + 5 x - 6

3 x^{2} + 5 = - 8 x

ln (1) + 1

7^{2} + 4^{2} = x^{2}

- 2 x - 4 = 2

What are the solutions to the equation 0^2+y^2+4(0)-10y+4=0 ?
The solutions to the equation 0^2+y^2+4(0)-10y+4=0 are y=5+sqrt(21),y=5-sqrt(21)
Find the zeros of 0^2+y^2+4(0)-10y+4=0
The zeros of 0^2+y^2+4(0)-10y+4=0 are y=5+sqrt(21),y=5-sqrt(21)