What are the solutions to the equation 5y^2-3y-5=0 ?

The solutions to the equation 5y^2-3y-5=0 are y=(3+sqrt(109))/(10),y=(3-sqrt(109))/(10)

Find the zeros of 5y^2-3y-5=0

The zeros of 5y^2-3y-5=0 are y=(3+sqrt(109))/(10),y=(3-sqrt(109))/(10)

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5 y^{2} - 3 y - 5 = 0

y = \frac{3 + 109}{10}, y = \frac{3 - 109}{10}

Decimal Notation

y = 1.34403 \dots, y = - 0.74403 \dots

Solution steps

5 y^{2} - 3 y - 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

(- 3)^{2} - 4 \cdot 5 (- 5) = 109

y_{1, 2} = \frac{- ( - 3 ) \pm 109}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 109}{2 \cdot 5}, y_{2} = \frac{- ( - 3 ) - 109}{2 \cdot 5}

y = \frac{- ( - 3 ) + 109}{2 \cdot 5} : \frac{3 + 109}{10}

y = \frac{- ( - 3 ) - 109}{2 \cdot 5} : \frac{3 - 109}{10}

The solutions to the quadratic equation are:

y = \frac{3 + 109}{10}, y = \frac{3 - 109}{10}

2 y - 10 + x y - 5 x

\frac{x + 2}{x + 3} < \frac{x - 1}{x - 2}

n n^{2} - 1

6^{2} + 4^{2} = c^{2}

2 x^{2} + 7 x \leq 15

What are the solutions to the equation 5y^2-3y-5=0 ?
The solutions to the equation 5y^2-3y-5=0 are y=(3+sqrt(109))/(10),y=(3-sqrt(109))/(10)
Find the zeros of 5y^2-3y-5=0
The zeros of 5y^2-3y-5=0 are y=(3+sqrt(109))/(10),y=(3-sqrt(109))/(10)