What are the solutions to the equation 3y^2-24y-47=0 ?

The solutions to the equation 3y^2-24y-47=0 are y=(12+sqrt(285))/3 ,y=(12-sqrt(285))/3

Find the zeros of 3y^2-24y-47=0

The zeros of 3y^2-24y-47=0 are y=(12+sqrt(285))/3 ,y=(12-sqrt(285))/3

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3 y^{2} - 24 y - 47 = 0

y = \frac{12 + 285}{3}, y = \frac{12 - 285}{3}

Decimal Notation

y = 9.62731 \dots, y = - 1.62731 \dots

Solution steps

3 y^{2} - 24 y - 47 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 3 ( - 47 )}{2 \cdot 3}

(- 24)^{2} - 4 \cdot 3 (- 47) = 2285

y_{1, 2} = \frac{- ( - 24 ) \pm 2 285}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 24 ) + 2 285}{2 \cdot 3}, y_{2} = \frac{- ( - 24 ) - 2 285}{2 \cdot 3}

y = \frac{- ( - 24 ) + 2 285}{2 \cdot 3} : \frac{12 + 285}{3}

y = \frac{- ( - 24 ) - 2 285}{2 \cdot 3} : \frac{12 - 285}{3}

The solutions to the quadratic equation are:

y = \frac{12 + 285}{3}, y = \frac{12 - 285}{3}

8 x^{2} - 5 x - 7 = 0

x^{2} - 4 x + 3

y^{2} - 4 y - 8 = 0

b, b^{2} - m^{2} = 0

4 x^{2} + 53 = 28 x

What are the solutions to the equation 3y^2-24y-47=0 ?
The solutions to the equation 3y^2-24y-47=0 are y=(12+sqrt(285))/3 ,y=(12-sqrt(285))/3
Find the zeros of 3y^2-24y-47=0
The zeros of 3y^2-24y-47=0 are y=(12+sqrt(285))/3 ,y=(12-sqrt(285))/3