Question 87904
Let's use the quadratic formula to solve for x:

Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general form of the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{-x^2+3*x+3=0}}} (notice {{{a=-1}}}, {{{b=3}}}, and {{{c=3}}})


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




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




{{{x = (-3 +- sqrt( 9+12 ))/(2*-1)}}} Multiply {{{-4*3*-1}}} to get {{{12}}}




{{{x = (-3 +- sqrt( 21 ))/(2*-1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-3 +- sqrt(21))/(2*-1)}}} Simplify the square root




{{{x = (-3 +- sqrt(21))/-2}}} Multiply 2 and -1 to get -2


So now the expression breaks down into two parts


{{{x = (-3 + sqrt(21))/-2}}} or {{{x = (-3 - sqrt(21))/-2}}}



Which approximate to


{{{x=-0.79128784747792}}} or {{{x=3.79128784747792}}}



So our solutions are:

{{{x=-0.79128784747792}}} or {{{x=3.79128784747792}}}


Notice when we graph {{{-x^2+3*x+3}}} we get:


{{{ graph( 500, 500, -10.7912878474779, 13.7912878474779, -10.7912878474779, 13.7912878474779,-1*x^2+3*x+3) }}}


when we use the root finder feature on a calculator, we find that {{{x=-0.79128784747792}}} and {{{x=3.79128784747792}}}.So this verifies our answer