Question 86176
Starting with the general quadratic


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


the general form of the quadratic equation is:


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


So lets solve {{{3*x^2+2*x-4}}}


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




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




{{{x = (-2 +- sqrt( 4+48 ))/(2*3)}}} Multiply {{{-4*-4*3}}} to get {{{48}}}




{{{x = (-2 +- sqrt( 52 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-2 +- 2*sqrt(13))/(2*3)}}} Simplify the square root




{{{x = (-2 +- 2*sqrt(13))/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{x = (-2 + 2*sqrt(13))/6}}} or {{{x = (-2 - 2*sqrt(13))/6}}}



Which approximate to


{{{x=0.86851709182133}}} or {{{x=-1.535183758488}}}



So our solutions are:

{{{x=0.86851709182133}}} or {{{x=-1.535183758488}}}


Notice when we graph {{{3*x^2+2*x-4}}} we get:


{{{ graph( 500, 500, -11.535183758488, 10.8685170918213, -11.535183758488, 10.8685170918213,3*x^2+2*x+-4) }}}


when we use the root finder feature on our calculator, we find that {{{x=0.86851709182133}}} and {{{x=-1.535183758488}}}.So this verifies our answer