@boxcy_: This formula is a way to find the solutions (roots) of a quadratic equation. Normally, when solving a quadratic, we use a formula that involves adding or subtracting the square root of a certain combination of the equation’s numbers. This version is just a rearranged form of the usual solution. Instead of dividing by twice the first number of the equation, it divides twice the last number of the equation by a combination that involves the middle number and the square root. This rearrangement is often used in computers or calculators because it avoids problems that happen when subtracting two very large numbers that are almost equal. It gives a more accurate result in such cases. In short: it finds the same solutions as the standard formula but in a way that can reduce rounding errors. #math #mathematics #study #studytok #quadratic

sagi :

c=0 then the 2nd formula is fvcked

Finding my dream 💖🫂 :

what is that 😭

Alkaid :

Че за формула и для чего

grapedafruit :

by the same logic then x=c/a ????

parii :

i do it only is a=0

*Dovy* :

what if c turns out to be 0 😭😭

leannneeeenx :

bro what really??? there's smth like that

ILLKA|_228 :

а если с=0

bshd78e8h :

nej paro 1428 :

I don't know

Muhammad Nizam :

Kenapa rumus ini tidak dijelaskan dalam buku kurikulum merdeka? lalu kenapa rumus ini jarang dipakai meski memiliki tingkat error yang lebih minim dibanding rumus umum?

AEB :

а это работает

Нету :

я впервые вижу эту формулу... Вау...

𝓐𝓶𝓲𝓻 :

قانون العام

Quitly :

@Boxcycratic equation

Ahmed abdellaoui :

😳😳😳

probability :

🥰🥰🥰

pomidor :

🤣

mohamed🇪🇬☪️(🇵🇸🇵🇸🇵🇸) :

😳😳😳

krillin :

😂😂😂

матвей :

😱😱😱

Hamza Walid :

(-b±√[b²-4ac])/2a let b²-4ac=D so it becomes (-b±√D)/2a rationalise the numerator so multiply up and down by its conjugate (-b∓√D) so it becomes (b²-D)/2a(-b∓√D) substitute D, 4ac/2a(-b∓√[b²-4ac]) ∴=2c/(-b∓√[b²-4ac]) this is bot popular because we usually rationalise the denominator not the not the numerator

Thalles Motta :

√B2+4ac/2a= b²+4d/2a 2c/2a c¹

coretradingmastery :

If it's a function with a y intercept at the origin then this gets cooked because it'll have no roots.