Question 118217


{{{3x^2-2x=15x-10}}} Start with the given equation



{{{3x^2-2x-15x+10=0}}}  Subtract 15x from both sides.  Add 10 to both sides. 




{{{3x^2-17x+10=0}}} Combine like terms



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^2-17*x+10=0}}} ( notice {{{a=3}}}, {{{b=-17}}}, and {{{c=10}}})





{{{x = (--17 +- sqrt( (-17)^2-4*3*10 ))/(2*3)}}} Plug in a=3, b=-17, and c=10




{{{x = (17 +- sqrt( (-17)^2-4*3*10 ))/(2*3)}}} Negate -17 to get 17




{{{x = (17 +- sqrt( 289-4*3*10 ))/(2*3)}}} Square -17 to get 289  (note: remember when you square -17, you must square the negative as well. This is because {{{(-17)^2=-17*-17=289}}}.)




{{{x = (17 +- sqrt( 289+-120 ))/(2*3)}}} Multiply {{{-4*10*3}}} to get {{{-120}}}




{{{x = (17 +- sqrt( 169 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (17 +- 13)/(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 = (17 +- 13)/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{x = (17 + 13)/6}}} or {{{x = (17 - 13)/6}}}


Lets look at the first part:


{{{x=(17 + 13)/6}}}


{{{x=30/6}}} Add the terms in the numerator

{{{x=5}}} Divide


So one answer is

{{{x=5}}}




Now lets look at the second part:


{{{x=(17 - 13)/6}}}


{{{x=4/6}}} Subtract the terms in the numerator

{{{x=2/3}}} Divide


So another answer is

{{{x=2/3}}}


So our solutions are:

{{{x=5}}} or {{{x=2/3}}}


Notice when we graph {{{3*x^2-17*x+10}}}, we get:


{{{ graph( 500, 500, -8, 15, -8, 15,3*x^2+-17*x+10) }}}


and we can see that the roots are {{{x=5}}} and {{{x=2/3}}}. This verifies our answer