What are the solutions to the equation 25y^2-9y-180=0 ?

The solutions to the equation 25y^2-9y-180=0 are y=(9+21sqrt(41))/(50),y=(9-21sqrt(41))/(50)

Find the zeros of 25y^2-9y-180=0

The zeros of 25y^2-9y-180=0 are y=(9+21sqrt(41))/(50),y=(9-21sqrt(41))/(50)

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25 y^{2} - 9 y - 180 = 0

y = \frac{9 + 21 41}{50}, y = \frac{9 - 21 41}{50}

Decimal Notation

y = 2.86931 \dots, y = - 2.50931 \dots

Solution steps

25 y^{2} - 9 y - 180 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 25 ( - 180 )}{2 \cdot 25}

(- 9)^{2} - 4 \cdot 25 (- 180) = 2141

y_{1, 2} = \frac{- ( - 9 ) \pm 21 41}{2 \cdot 25}

Separate the solutions

y_{1} = \frac{- ( - 9 ) + 21 41}{2 \cdot 25}, y_{2} = \frac{- ( - 9 ) - 21 41}{2 \cdot 25}

y = \frac{- ( - 9 ) + 21 41}{2 \cdot 25} : \frac{9 + 21 41}{50}

y = \frac{- ( - 9 ) - 21 41}{2 \cdot 25} : \frac{9 - 21 41}{50}

The solutions to the quadratic equation are:

y = \frac{9 + 21 41}{50}, y = \frac{9 - 21 41}{50}

5 x^{2} - 31 x - 72 = 0

80 + 40 x - 4.9 x^{2} = 0

x^{2} + 16 = 12 x

5 t^{2} = 30 t - 40

x, x^{2} y = a^{3}

What are the solutions to the equation 25y^2-9y-180=0 ?
The solutions to the equation 25y^2-9y-180=0 are y=(9+21sqrt(41))/(50),y=(9-21sqrt(41))/(50)
Find the zeros of 25y^2-9y-180=0
The zeros of 25y^2-9y-180=0 are y=(9+21sqrt(41))/(50),y=(9-21sqrt(41))/(50)