Question 125908
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 {{{x^2+5*x-14=0}}} ( notice {{{a=1}}}, {{{b=5}}}, and {{{c=-14}}})





{{{x = (-5 +- sqrt( (5)^2-4*1*-14 ))/(2*1)}}} Plug in a=1, b=5, and c=-14




{{{x = (-5 +- sqrt( 25-4*1*-14 ))/(2*1)}}} Square 5 to get 25  




{{{x = (-5 +- sqrt( 25+56 ))/(2*1)}}} Multiply {{{-4*-14*1}}} to get {{{56}}}




{{{x = (-5 +- sqrt( 81 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-5 +- 9)/(2*1)}}} 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 = (-5 +- 9)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-5 + 9)/2}}} or {{{x = (-5 - 9)/2}}}


Lets look at the first part:


{{{x=(-5 + 9)/2}}}


{{{x=4/2}}} Add the terms in the numerator

{{{x=2}}} Divide


So one answer is

{{{x=2}}}




Now lets look at the second part:


{{{x=(-5 - 9)/2}}}


{{{x=-14/2}}} Subtract the terms in the numerator

{{{x=-7}}} Divide


So another answer is

{{{x=-7}}}


So our solutions are:

{{{x=2}}} or {{{x=-7}}}


Notice when we graph {{{x^2+5*x-14}}}, we get:


{{{ graph( 500, 500, -17, 12, -17, 12,1*x^2+5*x+-14) }}}


and we can see that the roots are {{{x=2}}} and {{{x=-7}}}. This verifies our answer