Question 91863
{{{12x^2 + 7x =10}}} I'm assuming it looks like this right?


{{{12x^2 + 7x -10=0}}}  Subtract 10 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 {{{12*x^2+7*x-10=0}}} ( notice {{{a=12}}}, {{{b=7}}}, and {{{c=-10}}})


{{{x = (-7 +- sqrt( (7)^2-4*12*-10 ))/(2*12)}}} Plug in a=12, b=7, and c=-10




{{{x = (-7 +- sqrt( 49-4*12*-10 ))/(2*12)}}} Square 7 to get 49  




{{{x = (-7 +- sqrt( 49+480 ))/(2*12)}}} Multiply {{{-4*-10*12}}} to get {{{480}}}




{{{x = (-7 +- sqrt( 529 ))/(2*12)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-7 +- 23)/(2*12)}}} 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 = (-7 +- 23)/24}}} Multiply 2 and 12 to get 24


So now the expression breaks down into two parts


{{{x = (-7 + 23)/24}}} or {{{x = (-7 - 23)/24}}}


Lets look at the first part:


{{{x=(-7 + 23)/24}}}


{{{x=16/24}}} Add the terms in the numerator

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


So one answer is

{{{x=2/3}}}




Now lets look at the second part:


{{{x=(-7 - 23)/24}}}


{{{x=-30/24}}} Subtract the terms in the numerator

{{{x=-5/4}}} Divide


So another answer is

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


So our solutions are:

{{{x=2/3}}} or {{{x=-5/4}}}


Notice when we graph {{{12*x^2+7*x-10}}}, we get:


{{{ graph( 500, 500, -15, 12, -15, 12,12*x^2+7*x+-10) }}}


and we can see that the roots are {{{x=2/3}}} and {{{x=-5/4}}}. This verifies our answer