What are the solutions to the equation 3y^2+6y+3=0 ?

The solutions to the equation 3y^2+6y+3=0 are y=-1

Find the zeros of 3y^2+6y+3=0

The zeros of 3y^2+6y+3=0 are y=-1

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3 y^{2} + 6 y + 3 = 0

y = - 1

Solution steps

3 y^{2} + 6 y + 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

6^{2} - 4 \cdot 3 \cdot 3 = 0

y_{1, 2} = \frac{- 6 \pm 0}{2 \cdot 3}

y = \frac{- 6}{2 \cdot 3}

\frac{- 6}{2 \cdot 3} = - 1

y = - 1

The solution to the quadratic equation is:

y = - 1

\frac{\frac{8}{17}}{\frac{15}{17}}

\frac{3 x + 2}{2} = \frac{2 x - 3}{3}

\frac{2}{3} x - 3 = \frac{1}{2} x - 1

\frac{15 - 11 x}{x ^{2} - 6 x + 9} \leq 1

n^{2} - 9 n + 8 = 0

What are the solutions to the equation 3y^2+6y+3=0 ?
The solutions to the equation 3y^2+6y+3=0 are y=-1
Find the zeros of 3y^2+6y+3=0
The zeros of 3y^2+6y+3=0 are y=-1