What are the solutions to the equation 5y^2+10y+4=0 ?

The solutions to the equation 5y^2+10y+4=0 are y=(-5+sqrt(5))/5 ,y=-(sqrt(5)+1)/(sqrt(5))

Find the zeros of 5y^2+10y+4=0

The zeros of 5y^2+10y+4=0 are y=(-5+sqrt(5))/5 ,y=-(sqrt(5)+1)/(sqrt(5))

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5 y^{2} + 10 y + 4 = 0

y = \frac{- 5 + 5}{5}, y = - \frac{5 + 1}{5}

Decimal Notation

y = - 0.5527864045, y = - 1.44721 \dots

Solution steps

5 y^{2} + 10 y + 4 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 5 \cdot 4}{2 \cdot 5}

1 0^{2} - 4 \cdot 5 \cdot 4 = 25

y_{1, 2} = \frac{- 10 \pm 2 5}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- 10 + 2 5}{2 \cdot 5}, y_{2} = \frac{- 10 - 2 5}{2 \cdot 5}

y = \frac{- 10 + 2 5}{2 \cdot 5} : \frac{- 5 + 5}{5}

y = \frac{- 10 - 2 5}{2 \cdot 5} : - \frac{5 + 1}{5}

The solutions to the quadratic equation are:

y = \frac{- 5 + 5}{5}, y = - \frac{5 + 1}{5}

x^{2} + (- 2 x + c)^{2} - 10 x - 8 (- 2 x + c) + 21 = 0

- 5 y - 3 = 2 y^{2}

- 45 = 0 + 10 t + \frac{1}{2} (- 0.5) t^{2}

x^{2} - 0.57 \cdot 1 0^{- 10} + 3.8 \cdot 1 0^{- 10} \cdot x = 0

(x + 2)^{2} = 24

What are the solutions to the equation 5y^2+10y+4=0 ?
The solutions to the equation 5y^2+10y+4=0 are y=(-5+sqrt(5))/5 ,y=-(sqrt(5)+1)/(sqrt(5))
Find the zeros of 5y^2+10y+4=0
The zeros of 5y^2+10y+4=0 are y=(-5+sqrt(5))/5 ,y=-(sqrt(5)+1)/(sqrt(5))