What are the solutions to the equation 12x^2-180x+500=0 ?

The solutions to the equation 12x^2-180x+500=0 are x=(5(9+sqrt(21)))/6 ,x=(5(9-sqrt(21)))/6

Find the zeros of 12x^2-180x+500=0

The zeros of 12x^2-180x+500=0 are x=(5(9+sqrt(21)))/6 ,x=(5(9-sqrt(21)))/6

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12 x^{2} - 180 x + 500 = 0

x = \frac{5 ( 9 + 21 )}{6}, x = \frac{5 ( 9 - 21 )}{6}

Decimal Notation

x = 11.31881 \dots, x = 3.68118 \dots

Solution steps

12 x^{2} - 180 x + 500 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 180 ) \pm ( - 180 ) ^{2} - 4 \cdot 12 \cdot 500}{2 \cdot 12}

(- 180)^{2} - 4 \cdot 12 \cdot 500 = 2021

x_{1, 2} = \frac{- ( - 180 ) \pm 20 21}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 180 ) + 20 21}{2 \cdot 12}, x_{2} = \frac{- ( - 180 ) - 20 21}{2 \cdot 12}

x = \frac{- ( - 180 ) + 20 21}{2 \cdot 12} : \frac{5 ( 9 + 21 )}{6}

x = \frac{- ( - 180 ) - 20 21}{2 \cdot 12} : \frac{5 ( 9 - 21 )}{6}

The solutions to the quadratic equation are:

x = \frac{5 ( 9 + 21 )}{6}, x = \frac{5 ( 9 - 21 )}{6}

3 p^{2} + 4 p = 8

34 = 105 t - 16 t^{2}

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

z (z + 7) = 30

m^{2} - 8 m + 15 = 0

What are the solutions to the equation 12x^2-180x+500=0 ?
The solutions to the equation 12x^2-180x+500=0 are x=(5(9+sqrt(21)))/6 ,x=(5(9-sqrt(21)))/6
Find the zeros of 12x^2-180x+500=0
The zeros of 12x^2-180x+500=0 are x=(5(9+sqrt(21)))/6 ,x=(5(9-sqrt(21)))/6