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

The solutions to the equation y^2+3y-3=0 are y=(-3+sqrt(21))/2 ,y=(-3-sqrt(21))/2

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

The zeros of y^2+3y-3=0 are y=(-3+sqrt(21))/2 ,y=(-3-sqrt(21))/2

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

y = \frac{- 3 + 21}{2}, y = \frac{- 3 - 21}{2}

Decimal Notation

y = 0.79128 \dots, y = - 3.79128 \dots

Solution steps

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 3) = 21

y_{1, 2} = \frac{- 3 \pm 21}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 3 + 21}{2 \cdot 1}, y_{2} = \frac{- 3 - 21}{2 \cdot 1}

y = \frac{- 3 + 21}{2 \cdot 1} : \frac{- 3 + 21}{2}

y = \frac{- 3 - 21}{2 \cdot 1} : \frac{- 3 - 21}{2}

The solutions to the quadratic equation are:

y = \frac{- 3 + 21}{2}, y = \frac{- 3 - 21}{2}

15 - 0.06 x \geq 5

(x - 2) = x^{2} - 4

0 = 4 x^{2} + 2 x - 1

(x + 1)^{2} + 1 = 0

\frac{- 8 Q ^{2} + 20 Q + 800}{189} = 0

What are the solutions to the equation y^2+3y-3=0 ?
The solutions to the equation y^2+3y-3=0 are y=(-3+sqrt(21))/2 ,y=(-3-sqrt(21))/2
Find the zeros of y^2+3y-3=0
The zeros of y^2+3y-3=0 are y=(-3+sqrt(21))/2 ,y=(-3-sqrt(21))/2