Question 116773
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+4*x-15=0}}} ( notice {{{a=3}}}, {{{b=4}}}, and {{{c=-15}}})





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




{{{x = (-4 +- sqrt( 16-4*3*-15 ))/(2*3)}}} Square 4 to get 16  




{{{x = (-4 +- sqrt( 16+180 ))/(2*3)}}} Multiply {{{-4*-15*3}}} to get {{{180}}}




{{{x = (-4 +- sqrt( 196 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-4 +- 14)/(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 = (-4 +- 14)/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{x = (-4 + 14)/6}}} or {{{x = (-4 - 14)/6}}}


Lets look at the first part:


{{{x=(-4 + 14)/6}}}


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

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


So one answer is

{{{x=5/3}}}




Now lets look at the second part:


{{{x=(-4 - 14)/6}}}


{{{x=-18/6}}} Subtract the terms in the numerator

{{{x=-3}}} Divide


So another answer is

{{{x=-3}}}


So our solutions are:

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


Notice when we graph {{{3*x^2+4*x-15}}}, we get:


{{{ graph( 500, 500, -13, 15, -13, 15,3*x^2+4*x+-15) }}}


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