What are the solutions to the equation 12531r^2-4445060r+160070813=0 ?

The solutions to the equation 12531r^2-4445060r+160070813=0 are

Find the zeros of 12531r^2-4445060r+160070813=0

The zeros of 12531r^2-4445060r+160070813=0 are

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12531 r^{2} - 4445060 r + 160070813 = 0

r = \frac{4445060 + 12531 \frac{444506 0 ^{2}}{157025961} - \frac{640283252}{12531}}{25062}, r = \frac{4445060 - 12531 \frac{444506 0 ^{2}}{157025961} - \frac{640283252}{12531}}{25062}

Decimal Notation

r = 314.05008 \dots, r = 40.67499 \dots

Solution steps

12531 r^{2} - 4445060 r + 160070813 = 0

Divide both sides by

12531

\frac{12531 r ^{2}}{12531} - \frac{4445060 r}{12531} + \frac{160070813}{12531} = \frac{0}{12531}

Write in the standard form

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

r^{2} - \frac{4445060 r}{12531} + \frac{160070813}{12531} = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - \frac{4445060}{12531} ) \pm ( - \frac{4445060}{12531} ) ^{2} - 4 \cdot 1 \cdot \frac{160070813}{12531}}{2 \cdot 1}

(- \frac{4445060}{12531})^{2} - 4 \cdot 1 \cdot \frac{160070813}{12531} = \frac{444506 0 ^{2}}{157025961} - \frac{640283252}{12531}

r_{1, 2} = \frac{- ( - \frac{4445060}{12531} ) \pm \frac{444506 0 ^{2}}{157025961} - \frac{640283252}{12531}}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - \frac{4445060}{12531} ) + \frac{444506 0 ^{2}}{157025961} - \frac{640283252}{12531}}{2 \cdot 1}, r_{2} = \frac{- ( - \frac{4445060}{12531} ) - \frac{444506 0 ^{2}}{157025961} - \frac{640283252}{12531}}{2 \cdot 1}

r = \frac{- ( - \frac{4445060}{12531} ) + \frac{444506 0 ^{2}}{157025961} - \frac{640283252}{12531}}{2 \cdot 1} : \frac{4445060 + 12531 \frac{444506 0 ^{2}}{157025961} - \frac{640283252}{12531}}{25062}

r = \frac{- ( - \frac{4445060}{12531} ) - \frac{444506 0 ^{2}}{157025961} - \frac{640283252}{12531}}{2 \cdot 1} : \frac{4445060 - 12531 \frac{444506 0 ^{2}}{157025961} - \frac{640283252}{12531}}{25062}

The solutions to the quadratic equation are:

r = \frac{4445060 + 12531 \frac{444506 0 ^{2}}{157025961} - \frac{640283252}{12531}}{25062}, r = \frac{4445060 - 12531 \frac{444506 0 ^{2}}{157025961} - \frac{640283252}{12531}}{25062}

a^{2} - 10 a + 25 = 49

(\frac{- 5 b ^{3}}{a})^{3}

t^{2} + 24 t - 25

5 - x > 7

5 x - 9 + 3 (3 x + 3) = - 2 (x + 2)

What are the solutions to the equation 12531r^2-4445060r+160070813=0 ?
The solutions to the equation 12531r^2-4445060r+160070813=0 are
Find the zeros of 12531r^2-4445060r+160070813=0
The zeros of 12531r^2-4445060r+160070813=0 are