Question 124344
directions: use quadratic equation x=-b+(rad)(b^2-4ac)/2a to find theroots of the equation x^2-3x+2=0
:
The quadratic equation:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In your problem a=1; b=-3; c=2
{{{x = (-(-3) +- sqrt(-3^2 - 4 * 1 * 2 ))/(2*1) }}}
;
{{{x = (3 + sqrt(9 - 8 ))/2 }}}
{{{x = (3 + sqrt(1))/2}}}
{{{x = (3 + 1)/2}}}
{{{x = 4/2}}}
x = 2
and
{{{x = (3 - sqrt(9 - 8 ))/(2) }}}
{{{x = (3 - sqrt(1))/2}}}
{{{x = (3 - 1)/2}}}
{{{x = 2/2}}}
x = 1
:
The solutions are x = 2 and x = 1
I am sorry, I should not have assumed it was easy for you.