Question 90110
{{{4x^2=13x+12}}}


{{{0=-4x^2+13x+12}}} Subtract 4x^2 from both sides



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 {{{-4*x^2+13*x+12=0}}} ( notice {{{a=-4}}}, {{{b=13}}}, and {{{c=12}}})


{{{x = (-13 +- sqrt( (13)^2-4*-4*12 ))/(2*-4)}}} Plug in a=-4, b=13, and c=12




{{{x = (-13 +- sqrt( 169-4*-4*12 ))/(2*-4)}}} Square 13 to get 169  




{{{x = (-13 +- sqrt( 169+192 ))/(2*-4)}}} Multiply {{{-4*12*-4}}} to get {{{192}}}




{{{x = (-13 +- sqrt( 361 ))/(2*-4)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-13 +- 19)/(2*-4)}}} 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 = (-13 +- 19)/-8}}} Multiply 2 and -4 to get -8


So now the expression breaks down into two parts


{{{x = (-13 + 19)/-8}}} or {{{x = (-13 - 19)/-8}}}


Lets look at the first part:


{{{x=(-13 + 19)/-8}}}


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

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


So one answer is

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




Now lets look at the second part:


{{{x=(-13 - 19)/-8}}}


{{{x=-32/-8}}} Subtract the terms in the numerator

{{{x=4}}} Divide


So another answer is

{{{x=4}}}


So our solutions are:

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


Notice when we graph {{{-4*x^2+13*x+12}}}, we get:


{{{ graph( 500, 500, -13, 14, -13, 14,-4*x^2+13*x+12) }}}


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