Question 89386
Yes it's correct. You can use other methods such as the quadratic formula to verify your answer.


So 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+4*x-3=0}}} ( notice {{{a=1}}}, {{{b=4}}}, and {{{c=-3}}})


{{{x = (-4 +- sqrt( (4)^2-4*1*-3 ))/(2*1)}}} Plug in a=1, b=4, and c=-3




{{{x = (-4 +- sqrt( 16-4*1*-3 ))/(2*1)}}} Square 4 to get 16  




{{{x = (-4 +- sqrt( 16+12 ))/(2*1)}}} Multiply {{{-4*-3*1}}} to get {{{12}}}




{{{x = (-4 +- sqrt( 28 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




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


So now the expression breaks down into two parts


{{{x = (-4 + 2*sqrt(7))/2}}} or {{{x = (-4 - 2*sqrt(7))/2}}}



Now break up the fraction



{{{x=-4/2+2*sqrt(7)/2}}} or {{{x=-4/2-2*sqrt(7)/2}}}



Simplify



{{{x=-2+sqrt(7)}}} or {{{x=-2-sqrt(7)}}}


Notice we get the same answer. So our answer is verified.