What are the solutions to the equation 5y^2+9y-5=0 ?

The solutions to the equation 5y^2+9y-5=0 are y=(-9+sqrt(181))/(10),y=(-9-sqrt(181))/(10)

Find the zeros of 5y^2+9y-5=0

The zeros of 5y^2+9y-5=0 are y=(-9+sqrt(181))/(10),y=(-9-sqrt(181))/(10)

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5 y^{2} + 9 y - 5 = 0

y = \frac{- 9 + 181}{10}, y = \frac{- 9 - 181}{10}

Decimal Notation

y = 0.44536 \dots, y = - 2.24536 \dots

Solution steps

5 y^{2} + 9 y - 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

9^{2} - 4 \cdot 5 (- 5) = 181

y_{1, 2} = \frac{- 9 \pm 181}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- 9 + 181}{2 \cdot 5}, y_{2} = \frac{- 9 - 181}{2 \cdot 5}

y = \frac{- 9 + 181}{2 \cdot 5} : \frac{- 9 + 181}{10}

y = \frac{- 9 - 181}{2 \cdot 5} : \frac{- 9 - 181}{10}

The solutions to the quadratic equation are:

y = \frac{- 9 + 181}{10}, y = \frac{- 9 - 181}{10}

\frac{x ^{2}}{4 + 7} = 11

\frac{3}{5} x^{2} - x = \frac{1}{5}

x^{2} + 1 0^{2} = 1 3^{2}

(x^{2} - 2 x + 3) = 0

- x^{2} + 4 x - 2 = 0

What are the solutions to the equation 5y^2+9y-5=0 ?
The solutions to the equation 5y^2+9y-5=0 are y=(-9+sqrt(181))/(10),y=(-9-sqrt(181))/(10)
Find the zeros of 5y^2+9y-5=0
The zeros of 5y^2+9y-5=0 are y=(-9+sqrt(181))/(10),y=(-9-sqrt(181))/(10)