What are the solutions to the equation 1.2x^2+x=0 ?

The solutions to the equation 1.2x^2+x=0 are x=0,x=-5/6

Find the zeros of 1.2x^2+x=0

The zeros of 1.2x^2+x=0 are x=0,x=-5/6

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1.2 x^{2} + x = 0

x = 0, x = - \frac{5}{6}

Decimal Notation

x = 0, x = - 0.83333 \dots

Solution steps

1.2 x^{2} + x = 0

Multiply both sides by

10

12 x^{2} + 10 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 12 \cdot 0}{2 \cdot 12}

1 0^{2} - 4 \cdot 12 \cdot 0 = 10

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

Separate the solutions

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

x = \frac{- 10 + 10}{2 \cdot 12} : 0

x = \frac{- 10 - 10}{2 \cdot 12} : - \frac{5}{6}

The solutions to the quadratic equation are:

x = 0, x = - \frac{5}{6}

x^{2} = 6 x - 5

x^{2} - 2 x + 5 \leq 0

5 x^{2} - 125 = 0

- v^{2} + 9 = 0

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

What are the solutions to the equation 1.2x^2+x=0 ?
The solutions to the equation 1.2x^2+x=0 are x=0,x=-5/6
Find the zeros of 1.2x^2+x=0
The zeros of 1.2x^2+x=0 are x=0,x=-5/6