What are the solutions to the equation 6y^2+6y-12=0 ?

The solutions to the equation 6y^2+6y-12=0 are y=1,y=-2

Find the zeros of 6y^2+6y-12=0

The zeros of 6y^2+6y-12=0 are y=1,y=-2

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6 y^{2} + 6 y - 12 = 0

y = 1, y = - 2

Solution steps

6 y^{2} + 6 y - 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 6 ( - 12 )}{2 \cdot 6}

6^{2} - 4 \cdot 6 (- 12) = 18

y_{1, 2} = \frac{- 6 \pm 18}{2 \cdot 6}

Separate the solutions

y_{1} = \frac{- 6 + 18}{2 \cdot 6}, y_{2} = \frac{- 6 - 18}{2 \cdot 6}

y = \frac{- 6 + 18}{2 \cdot 6} : 1

y = \frac{- 6 - 18}{2 \cdot 6} : - 2

The solutions to the quadratic equation are:

y = 1, y = - 2

6 m^{4} + 27 m^{3} + 6 m^{2} + 27 m

\frac{n ( n + 1 ) ( 2 n + 1 )}{6} - \frac{n ( n + 1 )}{2}

- \frac{1}{5} \cdot 10

- 8 (n + 4) = - 32 - 8 n

2 b^{2} - b - 7 = 14

What are the solutions to the equation 6y^2+6y-12=0 ?
The solutions to the equation 6y^2+6y-12=0 are y=1,y=-2
Find the zeros of 6y^2+6y-12=0
The zeros of 6y^2+6y-12=0 are y=1,y=-2