Question 120458
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+12*x+37=0}}} ( notice {{{a=1}}}, {{{b=12}}}, and {{{c=37}}})





{{{x = (-12 +- sqrt( (12)^2-4*1*37 ))/(2*1)}}} Plug in a=1, b=12, and c=37




{{{x = (-12 +- sqrt( 144-4*1*37 ))/(2*1)}}} Square 12 to get 144  




{{{x = (-12 +- sqrt( 144+-148 ))/(2*1)}}} Multiply {{{-4*37*1}}} to get {{{-148}}}




{{{x = (-12 +- sqrt( -4 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




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




After simplifying, the quadratic has roots of


{{{x=-6 + i}}} or {{{x=-6 - i}}}