Question 187133
{{{f(x)= -x^2+3x+6}}} Start with the given function



{{{0=-x^2+3x+6}}} Plug in {{{f(x)=0}}}



{{{0=x^2-3x-6}}} Multiply EVERY term by -1 to make the leading coefficient positive.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-3}}}, and {{{c=-6}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-3) +- sqrt( (-3)^2-4(1)(-6) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-3}}}, and {{{c=-6}}}



{{{x = (3 +- sqrt( (-3)^2-4(1)(-6) ))/(2(1))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(1)(-6) ))/(2(1))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9--24 ))/(2(1))}}} Multiply {{{4(1)(-6)}}} to get {{{-24}}}



{{{x = (3 +- sqrt( 9+24 ))/(2(1))}}} Rewrite {{{sqrt(9--24)}}} as {{{sqrt(9+24)}}}



{{{x = (3 +- sqrt( 33 ))/(2(1))}}} Add {{{9}}} to {{{24}}} to get {{{33}}}



{{{x = (3 +- sqrt( 33 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (3+sqrt(33))/(2)}}} or {{{x = (3-sqrt(33))/(2)}}} Break up the expression.  



So the answers are {{{x = (3+sqrt(33))/(2)}}} or {{{x = (3-sqrt(33))/(2)}}} 



which approximate to {{{x=4.372}}} or {{{x=-1.372}}}