The best method of solving a quadratic. When your equation is in the standard form of ax² + bx + c = 0, x = (-b ± √b² - 4ac) / 2a. It is far superior to both factoring and completing the square. The equation may look difficult, but all you have to do is plug in the coefficients and simplify!

A: I had to use the quadratic formula to solve that tricky problem.

B: Why didn't you just do that in the first place?

B: Why didn't you just do that in the first place?

by Lord Quasar November 11, 2020

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A mildly long but sensible and useful equation. Typically used for solving for the two, one, or zero real values of X in a quadratic equation.

by マジで死にたい May 04, 2019

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x = -b plus or minus the square root of (b^2 - 4ac) all divided by 2a. Used for quadratic equations where you can't factor. the equation is ax^2 + bx + c.

by Yo fucking momma December 17, 2003

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A handy formula easily memorized by repetition. It is used in many instances, such as trying to find the roots of an equation, or solving an equation that will not factor with integers.

To solve using the Quadratic Formula:

In any equation

ax^2+bx+c,

The roots can be found by substituting into:

-b(+,-)sqrt(b^2-4ac)/2a

OR

Negative "B" plus or minus the square root of "B" squared minus 4(a)(c), all divided by 2(a).

To solve using the Quadratic Formula:

In any equation

ax^2+bx+c,

The roots can be found by substituting into:

-b(+,-)sqrt(b^2-4ac)/2a

OR

Negative "B" plus or minus the square root of "B" squared minus 4(a)(c), all divided by 2(a).

To solve using the Quadratic Formula:

In any equation

ax^2+bx+c,

The roots can be found by substituting into:

-b(+,-)sqrt(b^2-4ac)/2a

OR

Negative "B" plus or minus the square root of "B" squared minus 4(a)(c), all divided by 2(a).

EX:

Hm..

10x^2+13X-5

-13(+,-)sqrt(169+200)/20

>_>

In other words, by substituting the numbers in, you get 13(+,-)sqrt(369)/20.

As you may or may not have guessed, this gives you a very nasty number, which should be approximated and rounded to the nearest thousandth.

In any equation

ax^2+bx+c,

The roots can be found by substituting into:

-b(+,-)sqrt(b^2-4ac)/2a

OR

Negative "B" plus or minus the square root of "B" squared minus 4(a)(c), all divided by 2(a).

EX:

Hm..

10x^2+13X-5

-13(+,-)sqrt(169+200)/20

>_>

In other words, by substituting the numbers in, you get 13(+,-)sqrt(369)/20.

As you may or may not have guessed, this gives you a very nasty number, which should be approximated and rounded to the nearest thousandth.

by "Seagaia" May 19, 2005

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ex equals the opposite of B plus or minus the square root of B squared minus four AC all over two A (to the tune of the mulbary bush the monkey chases the weasle)

quadratic formula--->

x = -B(+/-)(squaroot of:)(b^2 - 4AC)

----------------------------------

2A

when the equation is

0=Ax^2 + Bx + C

x = -B(+/-)(squaroot of:)(b^2 - 4AC)

----------------------------------

2A

when the equation is

0=Ax^2 + Bx + C

by mathfreakazoid November 11, 2005

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The reason behind dropping out of chemistry. It causes ones brain to recoil in horror and claw at the inner skull wall...it proves that combining math and science is never a worthwhile endeaver...especially when you can just as easily become a biologist and cut up weird things for a living....much more enjoyable.

to hell with the quadratic formula

by Petsche May 15, 2005

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