Question 112065
{{{5x^2-12=-17x}}} Start with the given equation



{{{5x^2-12+17x=0}}} Add 17x to both sides



{{{5x^2+17x-12=0}}} Rearrange the 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 {{{5*x^2+17*x-12=0}}} ( notice {{{a=5}}}, {{{b=17}}}, and {{{c=-12}}})





{{{x = (-17 +- sqrt( (17)^2-4*5*-12 ))/(2*5)}}} Plug in a=5, b=17, and c=-12




{{{x = (-17 +- sqrt( 289-4*5*-12 ))/(2*5)}}} Square 17 to get 289  




{{{x = (-17 +- sqrt( 289+240 ))/(2*5)}}} Multiply {{{-4*-12*5}}} to get {{{240}}}




{{{x = (-17 +- sqrt( 529 ))/(2*5)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-17 +- 23)/(2*5)}}} 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 +- 23)/10}}} Multiply 2 and 5 to get 10


So now the expression breaks down into two parts


{{{x = (-17 + 23)/10}}} or {{{x = (-17 - 23)/10}}}


Lets look at the first part:


{{{x=(-17 + 23)/10}}}


{{{x=6/10}}} Add the terms in the numerator

{{{x=3/5}}} Divide


So one answer is

{{{x=3/5}}}




Now lets look at the second part:


{{{x=(-17 - 23)/10}}}


{{{x=-40/10}}} Subtract the terms in the numerator

{{{x=-4}}} Divide


So another answer is

{{{x=-4}}}


So our solutions are:

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


Notice when we graph {{{5*x^2+17*x-12}}}, we get:


{{{ graph( 500, 500, -14, 13, -14, 13,5*x^2+17*x+-12) }}}


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