What are the solutions to the equation 3(1)^2+2(1)y+y^2=2 ?

The solutions to the equation 3(1)^2+2(1)y+y^2=2 are y=-1

Find the zeros of 3(1)^2+2(1)y+y^2=2

The zeros of 3(1)^2+2(1)y+y^2=2 are y=-1

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3 (1)^{2} + 2 (1) y + y^{2} = 2

y = - 1

Solution steps

3 (1)^{2} + 2 (1) y + y^{2} = 2

Expand

3 \cdot 1^{2} + 2 (1) y + y^{2} : 3 + 2 y + y^{2}

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

Move

2

to the left side

y^{2} + 2 y + 1 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot 1 = 0

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

y = \frac{- 2}{2 \cdot 1}

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

y = - 1

The solution to the quadratic equation is:

y = - 1

3 x^{2} - 12 = x^{2} + 38

x^{2} + 22 x + 28 = 28 x

6 x^{2} + 11 x - 10 = 0

36 x^{2} - 164 x + 192 = 0

2 x^{2} - 22 = x^{2} + 700

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