What are the solutions to the equation 9y^2-42y-245=0 ?

The solutions to the equation 9y^2-42y-245=0 are y=(7(1+sqrt(6)))/3 ,y=(7(1-sqrt(6)))/3

Find the zeros of 9y^2-42y-245=0

The zeros of 9y^2-42y-245=0 are y=(7(1+sqrt(6)))/3 ,y=(7(1-sqrt(6)))/3

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9 y^{2} - 42 y - 245 = 0

y = \frac{7 ( 1 + 6 )}{3}, y = \frac{7 ( 1 - 6 )}{3}

Decimal

y = 8.04880 \dots, y = - 3.38214 \dots

Solution steps

9 y^{2} - 42 y - 245 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 42 ) \pm ( - 42 ) ^{2} - 4 \cdot 9 ( - 245 )}{2 \cdot 9}

(- 42)^{2} - 4 \cdot 9 (- 245) = 426

y_{1, 2} = \frac{- ( - 42 ) \pm 42 6}{2 \cdot 9}

Separate the solutions

y_{1} = \frac{- ( - 42 ) + 42 6}{2 \cdot 9}, y_{2} = \frac{- ( - 42 ) - 42 6}{2 \cdot 9}

y = \frac{- ( - 42 ) + 42 6}{2 \cdot 9} : \frac{7 ( 1 + 6 )}{3}

y = \frac{- ( - 42 ) - 42 6}{2 \cdot 9} : \frac{7 ( 1 - 6 )}{3}

The solutions to the quadratic equation are:

y = \frac{7 ( 1 + 6 )}{3}, y = \frac{7 ( 1 - 6 )}{3}

x, - x^{2} + 4 y^{2} + 2 z^{2} = 8

- 2 x^{2} - 20 x + 1500 = 0

64 = 64 t - 16 t^{2}

- x (1 - x) - 6 = 0

- 6 x^{2} + 12 = 0

What are the solutions to the equation 9y^2-42y-245=0 ?
The solutions to the equation 9y^2-42y-245=0 are y=(7(1+sqrt(6)))/3 ,y=(7(1-sqrt(6)))/3
Find the zeros of 9y^2-42y-245=0
The zeros of 9y^2-42y-245=0 are y=(7(1+sqrt(6)))/3 ,y=(7(1-sqrt(6)))/3