Question 98239
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+11*x-20=0}}} ( notice {{{a=3}}}, {{{b=11}}}, and {{{c=-20}}})


{{{x = (-11 +- sqrt( (11)^2-4*3*-20 ))/(2*3)}}} Plug in a=3, b=11, and c=-20




{{{x = (-11 +- sqrt( 121-4*3*-20 ))/(2*3)}}} Square 11 to get 121  




{{{x = (-11 +- sqrt( 121+240 ))/(2*3)}}} Multiply {{{-4*-20*3}}} to get {{{240}}}




{{{x = (-11 +- sqrt( 361 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-11 +- 19)/(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 = (-11 +- 19)/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{x = (-11 + 19)/6}}} or {{{x = (-11 - 19)/6}}}


Lets look at the first part:


{{{x=(-11 + 19)/6}}}


{{{x=8/6}}} Add the terms in the numerator

{{{x=4/3}}} Divide


So one answer is

{{{x=4/3}}}




Now lets look at the second part:


{{{x=(-11 - 19)/6}}}


{{{x=-30/6}}} Subtract the terms in the numerator

{{{x=-5}}} Divide


So another answer is

{{{x=-5}}}


So our solutions are:

{{{x=4/3}}} or {{{x=-5}}}


Notice when we graph {{{3*x^2+11*x-20}}}, we get:


{{{ graph( 500, 500, -15, 14, -15, 14,3*x^2+11*x+-20) }}}


and we can see that the roots are {{{x=4/3}}} and {{{x=-5}}}. This verifies our answer