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docker41 [41]
3 years ago
5

How to solve this equation: ax² - 5x + 2 = 0

Mathematics
1 answer:
swat323 years ago
6 0

Answer:

x=\frac{5(+/-)\sqrt{25-8a}} {2a}

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

ax^{2} -5x+2=0

so

a=a\\b=-5\\c=2

substitute in the formula

x=\frac{5(+/-)\sqrt{-5^{2}-4(a)(2)}}{2a}

x=\frac{5(+/-)\sqrt{25-8a}} {2a}

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Answer: 300 yards

Step-by-step explanation:

Since b1 is equal to 14 yds and b2 is equal to 26 yds we can plug that into the equation below (in the image).

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Now that we have found (b1 + b2), you can multiply that by h (the height of the trapezoid) which is 15, so we get 40 x 15 = 600.

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300 yds is your final answer.

Hope this helps!

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Tcecarenko [31]

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