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

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

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

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

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

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

Decimal Notation

θ = - 0.004, θ = - 0.02

Solution steps

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

Find Least Common Multiplier of

12500, 125 : 12500

Multiply by LCM=

12500

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

Simplify

1 + 300 θ + 12500 θ^{2} = 0

Divide both sides by

12500

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

Write in the standard form

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

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

Solve with the quadratic formula

θ_{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}

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} = \frac{x ^{2}}{8}

5 = (2 - h)^{2} + (4 - 2 h)^{2}

x, (x - \frac{a}{2})^{2} + y^{2} = \frac{a ^{2}}{4}

144 = x^{2}

2 (3 x^{2} + 1) = 0

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