24-18\div2+(-9)^2

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

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

simplify solve for inverse tangent line

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−x²+x+12=0

Solution

x=−3, x=4

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Solve by:

Quadratic formula

Completing the square

Factoring method

One step at a time

−x²+x+12=0

Solve with the quadratic formula

x_{1, 2}=−1± √1²−4(−1)· 122(−1)

√1²−4(−1)· 12=7

x_{1, 2}=−1± 72(−1)

Separate the solutions

x₁=−1+72(−1) , x₂=−1−72(−1)

x=−1+72(−1) : −3

x=−1−72(−1) : 4

The solutions to the quadratic equation are:

x=−3, x=4

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

24-18\div2+(-9)^2

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