CPhill

(2x+1)*(6x-1)  =  12x^2 + 6x - 2x - 1   =  12x^2 + 4x - 1

We don't have enough info to solve this......your equation is correct, but s and a may - or may not - have integer solutons with just one equation..........we need to generate another equation  involving s and a, somehow

There's  nothing to "solve" ......we can "expand" this as......

(x - 4)^2  = (x - 4) ( x - 4)  =  x^2 - 8x + 16

https://en.wikipedia.org/wiki/Schlieffen_Plan

Consider that  all the possible arrangements that can be made by choosing any 5 numbers from 97 = 97*96*95*94*93  = P(97,5)

Now ....consider all the possible sets that can be made by choosing any 5 cards from 97  =

C(97,5)........but .......since these sets can be arranged in any manner.....just order each one from the least number element to the greatest......and we will still have the same number of sets.....

So....the probability is given by :

[Number of ordered arrangements ]   /   [ Number of possible arrangements]  =

C(97,5)  / P(97/5)   =  (   97! /  [ (97 - 5)! *5! ] )  / [97! / (97 - 5)!]  =

[97! / 97!] * [ 97 - 5]!/ [97 - 5]! * [ 1 / 5! ]  = 1/120   =  about .833.....%

The hypotenuse is given by :

sqrt (15² + 20²) =   sqrt (225 + 400)  = sqrt(625)   =  25 cm

Another way this can be found directly is to notice that this is a multiple of a 3 - 4 - 5  right triangle

where  the two legs are 3 and 4.......and the hypotenuse is 5.........multiplying each of these numbers by 5 gives us   15 - 20 - 25........

3x^2-2x-2=4    add 2 to both sides

3x^2 - 2x    =   6        divide every term by 3

x^2 - (2/3)x  =   2      take 1/2 of 2/3, square it and add to both sides

x^2  - (2/3)x + 1/9   = 2 + 1/9        factor the left side

( x  - 1/3)^2  =  19/9         take the squareroot of both sides

x - 1/3   = pos/neg sqrt(19)/2       add 1/3    to both sides

x =     pos/neg sqrt(19)   + 1/3

Another way

 ([5 * 6 ] - 4 ) * 10  =

( 30  - 4 ) * 10  =

26 * 10   = 

260

2×{6+[12÷(3+1)]}-1  =

2*{6 +[ 12 / 4] } - 1  =

2* { 6 + 3}  - 1  =

2* {9}  - 1  =

18 - 1  =

17

It's actually an "old" avatar that I'm recycling........thanks, anyway.....LOL!!!!!