2 years ago

Mr. Heath organizes a six-game tournament where players will

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1 answer:

Svetradugi [14.3K]2 years ago

6 0

Answer:

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Step-by-step explanation:

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For this case, we have the following equation of the second degree:

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If we divide between 2 on both sides of the equation, we will have:

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Where:

$a = 15\\b = -14\\c = 3$

The solutions will come from:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

Substituting:

$x = \frac {- (- 14) \pm \sqrt {(- 14) ^ 2-4 (15) (3)}} {2 (15)}\\x = \frac {14 \pm \sqrt {196-180}} {30}\\x = \frac {14 \pm \sqrt {16}} {30}\\x = \frac {14 \pm4} {30}$

So, we have:

$x_ {1} = \frac{14 + 4} {30} = \frac {18} {30} = \frac {18} {30} = \frac {3} {5}\\x_ {2} = \frac {14-4} {30} = \frac {10} {30} = \frac {1} {3}\\$

Answer:

$x_ {1} = \frac {3} {5}\\x_ {2} = \frac {1} {3}$

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