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범위 2cos(2x-1)+4

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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x²+6x+3=19

Solution

x=2, x=−8

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x²+6x+3=19

3 를 오른쪽으로 이동

x²+6x=16

(x+a)²=b 양식으로 다시 쓰기

(x+3)²=25

위해서 f ²(x)=a 해결책은 f(x)=√a, −√a

x+3=√25 해결 : x=2

x+3=−√25 해결 : x=−8

2차 방정식의 해는 다음과 같다:

x=2, x=−8

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안녕, 질문하다 x²+6x+3=19 또는 아래 옵션 중에서 선택하세요

2차 방정식을 푸는 방법을 설명하세요

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

범위 2cos(2x-1)+4

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