Question 99342
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+2*x-35=0}}} ( notice {{{a=1}}}, {{{b=2}}}, and {{{c=-35}}})


{{{x = (-2 +- sqrt( (2)^2-4*1*-35 ))/(2*1)}}} Plug in a=1, b=2, and c=-35




{{{x = (-2 +- sqrt( 4-4*1*-35 ))/(2*1)}}} Square 2 to get 4  




{{{x = (-2 +- sqrt( 4+140 ))/(2*1)}}} Multiply {{{-4*-35*1}}} to get {{{140}}}




{{{x = (-2 +- sqrt( 144 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-2 +- 12)/(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 = (-2 +- 12)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-2 + 12)/2}}} or {{{x = (-2 - 12)/2}}}


Lets look at the first part:


{{{x=(-2 + 12)/2}}}


{{{x=10/2}}} Add the terms in the numerator

{{{x=5}}} Divide


So one answer is

{{{x=5}}}




Now lets look at the second part:


{{{x=(-2 - 12)/2}}}


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

{{{x=-7}}} Divide


So another answer is

{{{x=-7}}}


So our solutions are:

{{{x=5}}} or {{{x=-7}}}


Notice when we graph {{{x^2+2*x-35}}}, we get:


{{{ graph( 500, 500, -17, 15, -17, 15,1*x^2+2*x+-35) }}}


and we can see that the roots are {{{x=5}}} and {{{x=-7}}}. This verifies our answer