What are the solutions to the equation 20x^2+60x+27=0 ?

The solutions to the equation 20x^2+60x+27=0 are x=(3(sqrt(10)-5))/(10),x=-(3(5+sqrt(10)))/(10)

Find the zeros of 20x^2+60x+27=0

The zeros of 20x^2+60x+27=0 are x=(3(sqrt(10)-5))/(10),x=-(3(5+sqrt(10)))/(10)

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20 x^{2} + 60 x + 27 = 0

x = \frac{3 ( 10 - 5 )}{10}, x = - \frac{3 ( 5 + 10 )}{10}

Decimal Notation

x = - 0.55131 \dots, x = - 2.44868 \dots

Solution steps

20 x^{2} + 60 x + 27 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 60 \pm 6 0 ^{2} - 4 \cdot 20 \cdot 27}{2 \cdot 20}

6 0^{2} - 4 \cdot 20 \cdot 27 = 1210

x_{1, 2} = \frac{- 60 \pm 12 10}{2 \cdot 20}

Separate the solutions

x_{1} = \frac{- 60 + 12 10}{2 \cdot 20}, x_{2} = \frac{- 60 - 12 10}{2 \cdot 20}

x = \frac{- 60 + 12 10}{2 \cdot 20} : \frac{3 ( 10 - 5 )}{10}

x = \frac{- 60 - 12 10}{2 \cdot 20} : - \frac{3 ( 5 + 10 )}{10}

The solutions to the quadratic equation are:

x = \frac{3 ( 10 - 5 )}{10}, x = - \frac{3 ( 5 + 10 )}{10}

3.1416 > π

8 x^{3} - 18 x

6 x - 2 = - 4 x + 2

1.87 = 4.9 t^{2}

3 6^{- \frac{1}{2}}

What are the solutions to the equation 20x^2+60x+27=0 ?
The solutions to the equation 20x^2+60x+27=0 are x=(3(sqrt(10)-5))/(10),x=-(3(5+sqrt(10)))/(10)
Find the zeros of 20x^2+60x+27=0
The zeros of 20x^2+60x+27=0 are x=(3(sqrt(10)-5))/(10),x=-(3(5+sqrt(10)))/(10)