Question 102529
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-7=0}}} ( notice {{{a=1}}}, {{{b=5}}}, and {{{c=-7}}})





{{{x = (-5 +- sqrt( (5)^2-4*1*-7 ))/(2*1)}}} Plug in a=1, b=5, and c=-7




{{{x = (-5 +- sqrt( 25-4*1*-7 ))/(2*1)}}} Square 5 to get 25  




{{{x = (-5 +- sqrt( 25+28 ))/(2*1)}}} Multiply {{{-4*-7*1}}} to get {{{28}}}




{{{x = (-5 +- sqrt( 53 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-5 +- sqrt(53))/(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 +- sqrt(53))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-5 + sqrt(53))/2}}} or {{{x = (-5 - sqrt(53))/2}}}



Now break up the fraction



{{{x=-5/2+sqrt(53)/2}}} or {{{x=-5/2-sqrt(53)/2}}}



Simplify



{{{x=-5 / 2+sqrt(53)/2}}} or {{{x=-5 / 2-sqrt(53)/2}}}



So these expressions approximate to


{{{x=1.14005494464026}}} or {{{x=-6.14005494464026}}}



So our solutions are:

{{{x=1.14005494464026}}} or {{{x=-6.14005494464026}}}


Notice when we graph {{{x^2+5*x-7}}}, we get:


{{{ graph( 500, 500, -16.1400549446403, 11.1400549446403, -16.1400549446403, 11.1400549446403,1*x^2+5*x+-7) }}}


when we use the root finder feature on a calculator, we find that {{{x=1.14005494464026}}} and {{{x=-6.14005494464026}}}.So this verifies our answer