Question 111386
FOR {{{ax^2+bx+c=0}}}, {{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}


Here, {{{2x^2-3x-2=0 }}}, where {{{a=2}}}, {{{b=-3}}}, {{{c=-2}}}


Substituting,


{{{x=(-(3)+-sqrt((3)^2-(4*2*(-2)))))/(2*2)}}}
{{{x=(3+-sqrt(9-(-16)))/(4)}}}
{{{x=(3+-sqrt(9+16))/(4)}}}
{{{x=(3+-sqrt(25))/(4)}}}
{{{x=(3+-5)/(4)}}}

{{{x=(3+5)/4}}} or {{{x=(3-5)/4}}}
{{{x=8/4}}} or {{x=-2/4}}}
{{{x=2}}} or {{{x=-1/2}}}


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