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3 years ago

X²+8x-84=0 Formúla General

Mathematics

1 answer:

Brums [2.3K]3 years ago

8 0

Answer:

$x_{1} =-14\\\\x_{2} =6$

Step-by-step explanation:

$x^{2} +8x-84=0\\\\x=\dfrac{-b \pm \sqrt{b^{2}-4ac } }{2a} \\\\a=1; \: \: b=8; \: \: c=-84\\\\x=\dfrac{-8 \pm \sqrt{8^{2}-4 \cdot 1 \cdot (-84) } }{2 \cdot 1}=\dfrac{-8 \pm \sqrt{400 } }{2}=\dfrac{-8 \pm20}{2} =-4 \pm 10\\\\x_{1} =-14\\\\x_{2} =6$

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Two brothers want to share a large field equally.

Elan Coil [88]

Yes, the partition gives the two brothers equal shares.

Step-by-step explanation:

Step 1:

Assume the entire field has an area of B. So one brother takes $\frac{1}{12} B$ , $\frac{1}{6}B$ , and $\frac{1}{4} B$

So we need to calculate how much this brother takes in terms of B.

To do this we calculate how much $\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B$ is.

Step 2:

To add $\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B$ ,

First take the LCM of the denominators 12, 6, and 4

The LCM is 12, we multiply the denominator to get the LCM value, this same value is multiplied to the numerator too.

$\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B = \frac{1(1)}{12(1)} B + \frac{1(2)}{6(2)} B + \frac{1(3)}{4(3)} B \\\\\frac{1}{12} B + \frac{1}{6} B + \frac{1}{4} B= \frac{1B +2B+3B}{12} = \frac{B}{2}$