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inverse ofr ln(x)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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−x²+2x+8=0

Solution

x=−2, x=4

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Résoudre par la formule quadratique

Résoudre en complétant le carré

Résoudre en factorisant

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−x²+2x+8=0

Résoudre par la formule quadratique

x_{1, 2}=−2± √2²−4(−1)· 82(−1)

√2²−4(−1)· 8=6

x_{1, 2}=−2± 62(−1)

Séparer les solutions

x₁=−2+62(−1) , x₂=−2−62(−1)

x=−2+62(−1) : −2

x=−2−62(−1) : 4

Les solutions de l'équation de forme quadratique sont :

x=−2, x=4

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

inverse ofr ln(x)

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