Question 107039
You need to get them into the right form in
order to use the quadratic equation.. The right
form is {{{ax^2 + bx + c = 0}}}
{{{2x^2 - 5x = 3}}}
{{{2x^2 - 5x - 3 = 0}}}
a = 2
b = -5
c = -3
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (-(-5) +- sqrt( (-5)^2-4*2*(-3) ))/(2*2) }}}
{{{x = (5 +- sqrt(25 + 24 ))/4 }}}
{{{x = (5 +- sqrt(49 ))/4 }}}
{{{x = (5 + 7)/4}}}
{{{x = 3}}} answer
and
{{{x = (5 - 7)/4}}}
{{{-(1/2)}}} answer
You can substitute both answers back into the equation to check them
-------------------------------
{{{3x^2 - 2x + 1 = 0}}}
a = 3 
b = -2
c = 1
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (-(-2) +- sqrt( (-2)^2-4*3*1 ))/(2*3) }}}
{{{x = (2 +- sqrt( 4 - 12 ))/6 }}}
{{{x = (2 +- sqrt(-8))/6}}}
{{{x = (2 + 2*sqrt(2)i)/6}}}
{{{x = 1/3 + (sqrt(2)/3)i}}} answer
{{{x = (2 - 2*sqrt(2)i)/6}}}
{{{x = 1/3 - (sqrt(2)/3)i}}} answer
I checked the 2nd answer OK, It's tricky keeping
track of signs. Remember {{{i*i = -1}}}