Question 55251
Solve
{{{x^2/2 - 5x/4 = -3/4}}}  I hate fractions, don't you?  Let's obliterate them by multiplying both sides of the = sign by the LCD, 4.
{{{4(x^2/2)+4(-5x/4)=4(-3/4)}}}
{{{(4/2)x^2+(4/4)(-5x)=(4/4)(-3)}}}
{{{2x^2-5x=-3}}}  Set the problem = to 0.
{{{2x^2-5x+3=-3+3}}}
{{{2x^2-5x+3=0}}}  This quadratic equation is in standard form {{{highlight(ax^2+bx+c)}}} Factor, we need to find two integers that multiply to get a*c but add to get b.  Our a=2, b=-5, and c=3, a*c=6
-2*-3=6  and -2-3=-5
Replace b with -2-3 and factor by grouping:(there are other ways to do that, but this is easier to explain via the internet)
{{{2x^2+(-2-3)x+3=0}}}
{{{2x^2-2x-3x+3=0}}}
{{{(2x^2-2x)+(-3x+3)=0}}}
{{{2x(x-1)-3(x-1)=0}}}
{{{(2x-3)(x-1)=0}}}  Use the zero product property and set each parenthesis=0 and solve for x.
{{{2x-3=0}}} and {{{x-1=0}}}
{{{2x-3+3=0+3}}} and {{{x-1+1=0+1}}}
{{{2x=3}}} and {{{x=1}}}
{{{2x/2=3/2}}} and {{{x=1}}}
{{{x=3/2}}} and {{{x=1}}}
Some books write the solution set as x={1,3/2}
Happy Calculating!!!