Question 98784


    solve the equation for x

2(x-5)^2 =3

    2(x-5)^2 = 3

    2(x^2-10x+25) = 3

    2x^2-20x+50-3 = 0

    2x^2-20x+47 = 0

    compare with std form we get a= 2 , b= -20 and c = 47
         {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

      substituting for a,b and c we get , 

      x = -(-20)+-sq.rt (-20)^2-4.2.47/2.2

        = 20 +-sq.rt 400-376/4
        = 20+-sq.rt 24/4
        = 20+-2.sq.rt6/4
        = 2(10+-sq.rt 6)/4
        = (10+-sq.rt 6)/2