How do you solve $a^2-2a-224=0$ ?

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How do you solve $a^2-2a-224=0$ ?

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Answer 1 · 2015-07-08T05:05:07

Solve $y = x^2 - 2x - 224 = 0$

Explanation:

$y = x^2 - 2x - 224 = 0$ .
I use the new Transforming Method (Google Yahoo Search)
Find 2 numbers knowing sum (2) and product (-224). Roots have different signs.
Factor pairs of (-224) -> ...(-8, 28)(-14, 16). This sum is 2 = -b. Then, the 2 real roots are: $- 14 and 16.$

Answer 2 · 2015-07-08T06:26:28

Factor the trinomial as the product of two binomials. Set each binomial equal to zero, and solve for $a$ .

Explanation:

$a^2-2a-224=0$

Find two numbers that when added equal $-2$ , and when multiplied equal $-224$ . The numbers $-16$ and $14$ fit the pattern.

$a^2-2a-224=0$ =

$(a-16)(a+14)=0$

Set $(a-16)$ equal to zero and solve for $a$ .

$a-16=0$

#a=16

Set $(a+14)$ equal to zero and solve for $a$ .

$a+14=0$

$a=-14$

$a=16, -14$

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How do you solve $a^2-2a-224=0$ ?

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How do you solve a^2-2a-224=0a^2-2a-224=0?

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How do you solve $a^2-2a-224=0$ ?