Question 2756
 
    Before solving this problem, you must bear in your mind that the quadratic problems can be solved even by factorisation and not necessarily  by applying the formula i.e.{{{ax^2+bx+c}}} = 0 
 
    Choose 2 such numbers, the sum of which should be the coefficient of the mid-term of the expression and at the same time, the product of the same 2 numbers should be the last term multi-plied by coefficient of the 1st term.  I hope you understood this.  Here in your problem,    {{{10r^2 - 96r - 40}}}. The coefficient of the mid-term is -96 and the product of the coefficient of the first term and the last term is -400.
So we have to choose two such numbers as explained above.  So find out the factors of -400.
 
    I got -100 and +4.  It could be even 100 and -4; but the sum of these(+96) will not be equal to the coefficient of the  mid-term.  Isn't it? We will proceed further to solve your problem.  
 
    {{{10r^2 - 96r - 40}}}
 
 =  {{{10r^2 -100r + 4r - 40}}} ...  (-96r has been splitted as above)
 
 =  10r(r-10) + 4(r-10)   ...   (factorised the 1st and the last 2 terms)
 
 =  (r-10)*(10r+4)     ...   (Taken out the common factor and the other one)
 
     If (r-10)*(10r+4) = 0
 
    then, (r-10) or (10r+4) has to be = 0
 
    So r = 10;  or  r = -0.4     Answer.

I hope this is very clear to you now.