What are the solutions to the equation 12x^2-140x+300=0 ?

The solutions to the equation 12x^2-140x+300=0 are x=(5(7+sqrt(13)))/6 ,x=(5(7-sqrt(13)))/6

Find the zeros of 12x^2-140x+300=0

The zeros of 12x^2-140x+300=0 are x=(5(7+sqrt(13)))/6 ,x=(5(7-sqrt(13)))/6

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12 x^{2} - 140 x + 300 = 0

x = \frac{5 ( 7 + 13 )}{6}, x = \frac{5 ( 7 - 13 )}{6}

Decimal Notation

x = 8.83795 \dots, x = 2.82870 \dots

Solution steps

12 x^{2} - 140 x + 300 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 140 ) \pm ( - 140 ) ^{2} - 4 \cdot 12 \cdot 300}{2 \cdot 12}

(- 140)^{2} - 4 \cdot 12 \cdot 300 = 2013

x_{1, 2} = \frac{- ( - 140 ) \pm 20 13}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 140 ) + 20 13}{2 \cdot 12}, x_{2} = \frac{- ( - 140 ) - 20 13}{2 \cdot 12}

x = \frac{- ( - 140 ) + 20 13}{2 \cdot 12} : \frac{5 ( 7 + 13 )}{6}

x = \frac{- ( - 140 ) - 20 13}{2 \cdot 12} : \frac{5 ( 7 - 13 )}{6}

The solutions to the quadratic equation are:

x = \frac{5 ( 7 + 13 )}{6}, x = \frac{5 ( 7 - 13 )}{6}

2 x^{2} - 6 x + 3

2 x^{2} - 70 x + 480 = 0

4 y^{2} + 7 = 19

x^{2} + 8 x + \frac{112}{3} = 0

r^{2} + 80 r + 2000 = 0

What are the solutions to the equation 12x^2-140x+300=0 ?
The solutions to the equation 12x^2-140x+300=0 are x=(5(7+sqrt(13)))/6 ,x=(5(7-sqrt(13)))/6
Find the zeros of 12x^2-140x+300=0
The zeros of 12x^2-140x+300=0 are x=(5(7+sqrt(13)))/6 ,x=(5(7-sqrt(13)))/6