What are the solutions to the equation 24x^2-280x+600=0 ?

The solutions to the equation 24x^2-280x+600=0 are x=(5(7+sqrt(13)))/6 ,x=(5(7-sqrt(13)))/6

Find the zeros of 24x^2-280x+600=0

The zeros of 24x^2-280x+600=0 are x=(5(7+sqrt(13)))/6 ,x=(5(7-sqrt(13)))/6

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24 x^{2} - 280 x + 600 = 0

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

Solution steps

24 x^{2} - 280 x + 600 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 280 ) \pm ( - 280 ) ^{2} - 4 \cdot 24 \cdot 600}{2 \cdot 24}

(- 280)^{2} - 4 \cdot 24 \cdot 600 = 4013

x_{1, 2} = \frac{- ( - 280 ) \pm 40 13}{2 \cdot 24}

Separate the solutions

x_{1} = \frac{- ( - 280 ) + 40 13}{2 \cdot 24}, x_{2} = \frac{- ( - 280 ) - 40 13}{2 \cdot 24}

x = \frac{- ( - 280 ) + 40 13}{2 \cdot 24} : \frac{5 ( 7 + 13 )}{6}

x = \frac{- ( - 280 ) - 40 13}{2 \cdot 24} : \frac{5 ( 7 - 13 )}{6}

The solutions to the quadratic equation are:

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

x^{2} - (m^{2} - 5) x - 8 m + 3 = 0

6 + 4 x - x^{2}

\frac{x ^{2}}{3} = 27

7 x^{2} + 21 x + 14 = 0

2 (4 x)^{2} = 128

What are the solutions to the equation 24x^2-280x+600=0 ?
The solutions to the equation 24x^2-280x+600=0 are x=(5(7+sqrt(13)))/6 ,x=(5(7-sqrt(13)))/6
Find the zeros of 24x^2-280x+600=0
The zeros of 24x^2-280x+600=0 are x=(5(7+sqrt(13)))/6 ,x=(5(7-sqrt(13)))/6