geno3141

b² -4ac  is the discriminant

If  b² -4ac   >  0  then there are two real solutions.

If  b² -4ac   =  0  then there is one real solutions.

If  b² -4ac   <  0  then there are two imaginary/complex solutions.

(Technically, complex numbers include all real numbers as well as all imaginary numbers.)

-15 + n  =  -9

Add  15  to both sides:

-15 + 15 + n  =  -9 + 15

         n  =  6

You forgot to include the problem ...

Give us a particular type of problem, then we can help.

nx3 - 5  =  -4

First add 5 to both sides:

nx3 - 5 + 5   -4 + 5

nx3  =  1

Divide by 3:

nx3/3  =  1/3

n  =  1/3

m²  =  16/9

Find the square root of both sides; be sure to use the ± sign, because there will be two answers, one positive and one negative.

m  =  ±√(16/9)

m  =  ± √16 / √9

m  =  ± 4/3

365!  =  365 x 364 x 363 x ... x 332 x 331 x 330!

(365 - 35)!  =  330!

So:  365! / 330!  =  365 x 364 x 363 x ...x 332 x 331  (because the 330! in the numertor cancels the 330! in the denominator.

You are left with: [ 365 x 364 x 363 x ...x 332 x 331 ] / 365 ^ 35

Which would be:  365/365 x 364/365 x 363/365 x ... x 331/365

Now with patience, your answer will be ...

If you have lim(x→4) for some function, f(x)

and you want to change the limit to x→0, change the function  from f(x) to f(x-4) 

As an example:  lim(x→4) x² - 8   would change to:  lim(x→0) (x-4)² - 8 

2 236 000  -  17 740x  =  3 500 000

First, subtract 2 236 000 from both sides:

               - 17 740x  =  1 264 000

Now, divide both sides by -17 740

                            x  =  1 264 000 / -17 740  =  -71.3  (approx)

If you use combinations:  (2x + 5) ^ 5:

5nCr5·(2x)^5·(5)^0  +  5nCr4·(2x)^4·(5)^1  +  5nCr3·(2x)^3·(5)^2  +      5nCr2·(2x)^2·(5)^3  + 5nCr1·(2x)^1·(5)^4  +  5nCr0·(2x)^0·(5)^5

= 1·32x^5·1  + 5·16x^4·5  +  10·8x^3·25  +  10·4x^2·125  +  5·2x·625  +  1·x^0·3125

=  160x^5  +  400x^4  +  2000x^3  +  5000x^2  +  6250x + 3125