Question 102753
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^2-15*x+12=0}}} ( notice {{{a=3}}}, {{{b=-15}}}, and {{{c=12}}})





{{{x = (--15 +- sqrt( (-15)^2-4*3*12 ))/(2*3)}}} Plug in a=3, b=-15, and c=12




{{{x = (15 +- sqrt( (-15)^2-4*3*12 ))/(2*3)}}} Negate -15 to get 15




{{{x = (15 +- sqrt( 225-4*3*12 ))/(2*3)}}} Square -15 to get 225  (note: remember when you square -15, you must square the negative as well. This is because {{{(-15)^2=-15*-15=225}}}.)




{{{x = (15 +- sqrt( 225+-144 ))/(2*3)}}} Multiply {{{-4*12*3}}} to get {{{-144}}}




{{{x = (15 +- sqrt( 81 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (15 +- 9)/(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 = (15 +- 9)/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{x = (15 + 9)/6}}} or {{{x = (15 - 9)/6}}}


Lets look at the first part:


{{{x=(15 + 9)/6}}}


{{{x=24/6}}} Add the terms in the numerator

{{{x=4}}} Divide


So one answer is

{{{x=4}}}




Now lets look at the second part:


{{{x=(15 - 9)/6}}}


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

{{{x=1}}} Divide


So another answer is

{{{x=1}}}


So our solutions are:

{{{x=4}}} or {{{x=1}}}


Notice when we graph {{{3*x^2-15*x+12}}}, we get:


{{{ graph( 500, 500, -9, 14, -9, 14,3*x^2+-15*x+12) }}}


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