slope of x+5

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

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x²+10x−24=0

Solution

x=2, x=−12

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Résoudre par la formule quadratique

Résoudre en complétant le carré

Résoudre en factorisant

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x²+10x−24=0

Résoudre par la formule quadratique

x_{1, 2}=−10± √10²−4· 1· (−24)2· 1

√10²−4· 1· (−24)=14

x_{1, 2}=−10± 142· 1

Séparer les solutions

x₁=−10+142· 1 , x₂=−10−142· 1

x=−10+142· 1 : 2

x=−10−142· 1 : −12

Les solutions de l'équation de forme quadratique sont :

x=2, x=−12

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

slope of x+5

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