Question 122123
Given equation is 2x^2-5x=3
     2x^2-5x-3=0
     Now the coefficient of x^2 and constant multiplying, we get
     -6.
     Then, we find two numbers that numbers will be added we get -5(the   coefficient of x)and multiply that numbers will become -6.
     Therefore, that numbers are -6, 1
     Now the given equation becomes 2x^2-6x+x-3=0.
     (We take common number from first two variables.)
    i.e., 2x(x-3) + 1(x-3)=0
     i.e., (2x+1)=0,(x-3)=0 are the factors of the given eqn.
     Hence x=-1/2 and x=3.
                   (OR)
   The problem also solve by using the formula

http://www.algebra.com/cgi-bin/plot-formula.mpl?expression=x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29&x=0003

    In this equation a=2, b=-5, c=-3 and substitute these values in that formulae we will get x=-1/2 and x=3.