What are the solutions to the equation 1/12500+(3x)/(125)+x^2=0 ?

The solutions to the equation 1/12500+(3x)/(125)+x^2=0 are x=-1/250 ,x=-1/50

Find the zeros of 1/12500+(3x)/(125)+x^2=0

The zeros of 1/12500+(3x)/(125)+x^2=0 are x=-1/250 ,x=-1/50

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\frac{1}{12500} + \frac{3 x}{125} + x^{2} = 0

x = - \frac{1}{250}, x = - \frac{1}{50}

Decimal Notation

x = - 0.004, x = - 0.02

Solution steps

\frac{1}{12500} + \frac{3 x}{125} + x^{2} = 0

Find Least Common Multiplier of

12500, 125 : 12500

Multiply by LCM=

12500

\frac{1}{12500} \cdot 12500 + \frac{3 x}{125} \cdot 12500 + x^{2} \cdot 12500 = 0 \cdot 12500

Simplify

1 + 300 x + 12500 x^{2} = 0

Divide both sides by

12500

\frac{1}{12500} + \frac{300 x}{12500} + \frac{12500 x ^{2}}{12500} = \frac{0}{12500}

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + \frac{3 x}{125} + \frac{1}{12500} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- \frac{3}{125} \pm ( \frac{3}{125} ) ^{2} - 4 \cdot 1 \cdot \frac{1}{12500}}{2 \cdot 1}

(\frac{3}{125})^{2} - 4 \cdot 1 \cdot \frac{1}{12500} = \frac{2}{125}

x_{1, 2} = \frac{- \frac{3}{125} \pm \frac{2}{125}}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- \frac{3}{125} + \frac{2}{125}}{2 \cdot 1}, x_{2} = \frac{- \frac{3}{125} - \frac{2}{125}}{2 \cdot 1}

x = \frac{- \frac{3}{125} + \frac{2}{125}}{2 \cdot 1} : - \frac{1}{250}

x = \frac{- \frac{3}{125} - \frac{2}{125}}{2 \cdot 1} : - \frac{1}{50}

The solutions to the quadratic equation are:

x = - \frac{1}{250}, x = - \frac{1}{50}

x (x + 2) = 195

2 (x - 1) (2 x + 3) = 8 x - 5 x (2 - 2 x) - 10

(v + 5)^{2} = 5

3 x^{2} = - x + 10

3 x^{2} + 18 x - 21 = 0

What are the solutions to the equation 1/12500+(3x)/(125)+x^2=0 ?
The solutions to the equation 1/12500+(3x)/(125)+x^2=0 are x=-1/250 ,x=-1/50
Find the zeros of 1/12500+(3x)/(125)+x^2=0
The zeros of 1/12500+(3x)/(125)+x^2=0 are x=-1/250 ,x=-1/50