Question 94356
Let's use the quadratic formula to solve for x:



Starting with the general quadratic


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


the general solution using the quadratic equation is:


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


So lets solve {{{-3*x^2+6*x+23=0}}} ( notice {{{a=-3}}}, {{{b=6}}}, and {{{c=23}}})


{{{x = (-6 +- sqrt( (6)^2-4*-3*23 ))/(2*-3)}}} Plug in a=-3, b=6, and c=23




{{{x = (-6 +- sqrt( 36-4*-3*23 ))/(2*-3)}}} Square 6 to get 36  




{{{x = (-6 +- sqrt( 36+276 ))/(2*-3)}}} Multiply {{{-4*23*-3}}} to get {{{276}}}




{{{x = (-6 +- sqrt( 312 ))/(2*-3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-6 +- 2*sqrt(78))/(2*-3)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (-6 +- 2*sqrt(78))/-6}}} Multiply 2 and -3 to get -6


So now the expression breaks down into two parts


{{{x = (-6 + 2*sqrt(78))/-6}}} or {{{x = (-6 - 2*sqrt(78))/-6}}}



Now break up the fraction



{{{x=-6/-6+2*sqrt(78)/-6}}} or {{{x=-6/-6-2*sqrt(78)/-6}}}



Simplify



{{{x=1-sqrt(78)/3}}} or {{{x=1+sqrt(78)/3}}}



So these expressions approximate to


{{{x=-1.94392028877595}}} or {{{x=3.94392028877595}}}



So our solutions are:

{{{x=-1.94392028877595}}} or {{{x=3.94392028877595}}}


Notice when we graph {{{-3*x^2+6*x+23}}}, we get:


{{{ graph( 500, 500, -11.9439202887759, 13.9439202887759, -11.9439202887759, 13.9439202887759,-3*x^2+6*x+23) }}}


when we use the root finder feature on a calculator, we find that {{{x=-1.94392028877595}}} and {{{x=3.94392028877595}}}.So this verifies our answer