What are the solutions to the equation 209-120x+12x^2=0 ?

The solutions to the equation 209-120x+12x^2=0 are x=(30+sqrt(273))/6 ,x=(30-sqrt(273))/6

Find the zeros of 209-120x+12x^2=0

The zeros of 209-120x+12x^2=0 are x=(30+sqrt(273))/6 ,x=(30-sqrt(273))/6

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209 - 120 x + 12 x^{2} = 0

x = \frac{30 + 273}{6}, x = \frac{30 - 273}{6}

Decimal Notation

x = 7.75378 \dots, x = 2.24621 \dots

Solution steps

209 - 120 x + 12 x^{2} = 0

Write in the standard form

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

12 x^{2} - 120 x + 209 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 120 ) \pm ( - 120 ) ^{2} - 4 \cdot 12 \cdot 209}{2 \cdot 12}

(- 120)^{2} - 4 \cdot 12 \cdot 209 = 4273

x_{1, 2} = \frac{- ( - 120 ) \pm 4 273}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 120 ) + 4 273}{2 \cdot 12}, x_{2} = \frac{- ( - 120 ) - 4 273}{2 \cdot 12}

x = \frac{- ( - 120 ) + 4 273}{2 \cdot 12} : \frac{30 + 273}{6}

x = \frac{- ( - 120 ) - 4 273}{2 \cdot 12} : \frac{30 - 273}{6}

The solutions to the quadratic equation are:

x = \frac{30 + 273}{6}, x = \frac{30 - 273}{6}

x, x^{2} + y^{2} + 2 x + 1 = 0

x, 2 x^{2} - 4 x - 1 = 0

\frac{x ^{2} + 10}{2} = 19

x^{2} - 10 x - 4

2 x^{2} + 5 x + 3

What are the solutions to the equation 209-120x+12x^2=0 ?
The solutions to the equation 209-120x+12x^2=0 are x=(30+sqrt(273))/6 ,x=(30-sqrt(273))/6
Find the zeros of 209-120x+12x^2=0
The zeros of 209-120x+12x^2=0 are x=(30+sqrt(273))/6 ,x=(30-sqrt(273))/6