What are the solutions to the equation y^2+289/4 =-17y ?

The solutions to the equation y^2+289/4 =-17y are y=-17/2

Find the zeros of y^2+289/4 =-17y

The zeros of y^2+289/4 =-17y are y=-17/2

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y^{2} + \frac{289}{4} = - 17 y

y = - \frac{17}{2}

Decimal Notation

y = - 8.5

Solution steps

y^{2} + \frac{289}{4} = - 17 y

Multiply both sides by

4

4 y^{2} + 289 = - 68 y

Move

68 y

to the left side

4 y^{2} + 289 + 68 y = 0

Write in the standard form

a x^{2} + b x + c = 0

4 y^{2} + 68 y + 289 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 68 \pm 6 8 ^{2} - 4 \cdot 4 \cdot 289}{2 \cdot 4}

6 8^{2} - 4 \cdot 4 \cdot 289 = 0

y_{1, 2} = \frac{- 68 \pm 0}{2 \cdot 4}

y = \frac{- 68}{2 \cdot 4}

\frac{- 68}{2 \cdot 4} = - \frac{17}{2}

y = - \frac{17}{2}

The solution to the quadratic equation is:

y = - \frac{17}{2}

\frac{1}{2} (4) - 3

3 v - 5 = 40

9 a^{2} - 42 a + 49

7 (n + 6) > 14

\frac{3}{2} + 2 (x - 1) = 3 - 2 x

What are the solutions to the equation y^2+289/4 =-17y ?
The solutions to the equation y^2+289/4 =-17y are y=-17/2
Find the zeros of y^2+289/4 =-17y
The zeros of y^2+289/4 =-17y are y=-17/2