What are the solutions to the equation 3y(y-6)-8=12 ?

The solutions to the equation 3y(y-6)-8=12 are y=(9+sqrt(141))/3 ,y=(9-sqrt(141))/3

Find the zeros of 3y(y-6)-8=12

The zeros of 3y(y-6)-8=12 are y=(9+sqrt(141))/3 ,y=(9-sqrt(141))/3

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3 y (y - 6) - 8 = 12

y = \frac{9 + 141}{3}, y = \frac{9 - 141}{3}

Solution steps

3 y (y - 6) - 8 = 12

Expand

3 y (y - 6) - 8 : 3 y^{2} - 18 y - 8

3 y^{2} - 18 y - 8 = 12

Move

12

to the left side

3 y^{2} - 18 y - 20 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 3 ( - 20 )}{2 \cdot 3}

(- 18)^{2} - 4 \cdot 3 (- 20) = 2141

y_{1, 2} = \frac{- ( - 18 ) \pm 2 141}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 18 ) + 2 141}{2 \cdot 3}, y_{2} = \frac{- ( - 18 ) - 2 141}{2 \cdot 3}

y = \frac{- ( - 18 ) + 2 141}{2 \cdot 3} : \frac{9 + 141}{3}

y = \frac{- ( - 18 ) - 2 141}{2 \cdot 3} : \frac{9 - 141}{3}

The solutions to the quadratic equation are:

y = \frac{9 + 141}{3}, y = \frac{9 - 141}{3}

2 x^{2} - 1 = 3 x

8 x^{2} + 2 (x + 6) = 12

x^{2} + 22 x + 117 = 0

a, y^{2} + a^{2} = (a + y) 2 - a (a + 1)

x^{2} + 2 x = 15

What are the solutions to the equation 3y(y-6)-8=12 ?
The solutions to the equation 3y(y-6)-8=12 are y=(9+sqrt(141))/3 ,y=(9-sqrt(141))/3
Find the zeros of 3y(y-6)-8=12
The zeros of 3y(y-6)-8=12 are y=(9+sqrt(141))/3 ,y=(9-sqrt(141))/3