What are the solutions to the equation 1200-280x+12x^2=0 ?

The solutions to the equation 1200-280x+12x^2=0 are x=(5(7+sqrt(13)))/3 ,x=(5(7-sqrt(13)))/3

Find the zeros of 1200-280x+12x^2=0

The zeros of 1200-280x+12x^2=0 are x=(5(7+sqrt(13)))/3 ,x=(5(7-sqrt(13)))/3

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1200 - 280 x + 12 x^{2} = 0

x = \frac{5 ( 7 + 13 )}{3}, x = \frac{5 ( 7 - 13 )}{3}

Decimal Notation

x = 17.67591 \dots, x = 5.65741 \dots

Solution steps

1200 - 280 x + 12 x^{2} = 0

Write in the standard form

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

12 x^{2} - 280 x + 1200 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 280 ) \pm ( - 280 ) ^{2} - 4 \cdot 12 \cdot 1200}{2 \cdot 12}

(- 280)^{2} - 4 \cdot 12 \cdot 1200 = 4013

x_{1, 2} = \frac{- ( - 280 ) \pm 40 13}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 280 ) + 40 13}{2 \cdot 12}, x_{2} = \frac{- ( - 280 ) - 40 13}{2 \cdot 12}

x = \frac{- ( - 280 ) + 40 13}{2 \cdot 12} : \frac{5 ( 7 + 13 )}{3}

x = \frac{- ( - 280 ) - 40 13}{2 \cdot 12} : \frac{5 ( 7 - 13 )}{3}

The solutions to the quadratic equation are:

x = \frac{5 ( 7 + 13 )}{3}, x = \frac{5 ( 7 - 13 )}{3}

8 k^{2} = 544

3 x^{2} + 2 x - 8 = 0

12 x^{2} + 13 x - 35 = 0

a^{2} = - 1, 0 y (a) a^{10} = 24

0 = 3 x^{2} - 10 x + 4

What are the solutions to the equation 1200-280x+12x^2=0 ?
The solutions to the equation 1200-280x+12x^2=0 are x=(5(7+sqrt(13)))/3 ,x=(5(7-sqrt(13)))/3
Find the zeros of 1200-280x+12x^2=0
The zeros of 1200-280x+12x^2=0 are x=(5(7+sqrt(13)))/3 ,x=(5(7-sqrt(13)))/3