Question 118752
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+x-4=0}}} ( notice {{{a=3}}}, {{{b=1}}}, and {{{c=-4}}})





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




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




{{{x = (-1 +- sqrt( 1+48 ))/(2*3)}}} Multiply {{{-4*-4*3}}} to get {{{48}}}




{{{x = (-1 +- sqrt( 49 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-1 +- 7)/(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 = (-1 +- 7)/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{x = (-1 + 7)/6}}} or {{{x = (-1 - 7)/6}}}


Lets look at the first part:


{{{x=(-1 + 7)/6}}}


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

{{{x=1}}} Divide


So one answer is

{{{x=1}}}




Now lets look at the second part:


{{{x=(-1 - 7)/6}}}


{{{x=-8/6}}} Subtract the terms in the numerator

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


So another answer is

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


So our solutions are:

{{{x=1}}} or {{{x=-4/3}}}


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


{{{ graph( 500, 500, -14, 11, -14, 11,3*x^2+1*x+-4) }}}


and we can see that the roots are {{{x=1}}} and {{{x=-4/3}}}. This verifies our answer