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

The solutions to the equation y^2+3y=2 are y=(-3+sqrt(17))/2 ,y=(-3-sqrt(17))/2

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

The zeros of y^2+3y=2 are y=(-3+sqrt(17))/2 ,y=(-3-sqrt(17))/2

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y^{2} + 3 y = 2

y = \frac{- 3 + 17}{2}, y = \frac{- 3 - 17}{2}

Decimal Notation

y = 0.56155 \dots, y = - 3.56155 \dots

Solution steps

y^{2} + 3 y = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 2) = 17

y_{1, 2} = \frac{- 3 \pm 17}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 3 + 17}{2 \cdot 1}, y_{2} = \frac{- 3 - 17}{2 \cdot 1}

y = \frac{- 3 + 17}{2 \cdot 1} : \frac{- 3 + 17}{2}

y = \frac{- 3 - 17}{2 \cdot 1} : \frac{- 3 - 17}{2}

The solutions to the quadratic equation are:

y = \frac{- 3 + 17}{2}, y = \frac{- 3 - 17}{2}

(1 - x)^{2} = 9

12 x^{2} + 16 x - 3 = 0

x^{2} - 4 x = - 29

121 x^{2} - 8 = 0

- x^{2} + 4 x + 2

What are the solutions to the equation y^2+3y=2 ?
The solutions to the equation y^2+3y=2 are y=(-3+sqrt(17))/2 ,y=(-3-sqrt(17))/2
Find the zeros of y^2+3y=2
The zeros of y^2+3y=2 are y=(-3+sqrt(17))/2 ,y=(-3-sqrt(17))/2