What are the solutions to the equation 3y^2-24y-41=0 ?

The solutions to the equation 3y^2-24y-41=0 are y=(12+sqrt(267))/3 ,y=(12-sqrt(267))/3

Find the zeros of 3y^2-24y-41=0

The zeros of 3y^2-24y-41=0 are y=(12+sqrt(267))/3 ,y=(12-sqrt(267))/3

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3 y^{2} - 24 y - 41 = 0

y = \frac{12 + 267}{3}, y = \frac{12 - 267}{3}

Decimal Notation

y = 9.44671 \dots, y = - 1.44671 \dots

Solution steps

3 y^{2} - 24 y - 41 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 3 ( - 41 )}{2 \cdot 3}

(- 24)^{2} - 4 \cdot 3 (- 41) = 2267

y_{1, 2} = \frac{- ( - 24 ) \pm 2 267}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 24 ) + 2 267}{2 \cdot 3}, y_{2} = \frac{- ( - 24 ) - 2 267}{2 \cdot 3}

y = \frac{- ( - 24 ) + 2 267}{2 \cdot 3} : \frac{12 + 267}{3}

y = \frac{- ( - 24 ) - 2 267}{2 \cdot 3} : \frac{12 - 267}{3}

The solutions to the quadratic equation are:

y = \frac{12 + 267}{3}, y = \frac{12 - 267}{3}

(\frac{x}{4})^{2} - 0.16 = 0

x^{2} - 2 x - 5

3 x + 2 = - 2 x^{2} + 3 x + 6

6 p^{2} = 42

3 m^{2} + 15 m + 25.5 = 0

What are the solutions to the equation 3y^2-24y-41=0 ?
The solutions to the equation 3y^2-24y-41=0 are y=(12+sqrt(267))/3 ,y=(12-sqrt(267))/3
Find the zeros of 3y^2-24y-41=0
The zeros of 3y^2-24y-41=0 are y=(12+sqrt(267))/3 ,y=(12-sqrt(267))/3