Your polynomial is a quadratic, a second degree polynomial, to be exact. It can be factored to solve for those values of x you are looking for. Depending upon how you're factoring in class, solving can be done 1 of several ways but the never-fail way is the quadratic formula. Putting those values into the quadratic formula gives us this: [tex]x= \frac{-3+/- \sqrt{3^2-4(1)(5)} }{2(1)} [/tex] which simplifies to [tex]x= \frac{-3+/- \sqrt{-11} }{2} [/tex]. We have to deal with the square root of -11 now. Rewriting using the imaginary i gives us [tex]x= \frac{-3+/- \sqrt{-1(11)} }{2} [/tex]. Since -1 is equal to i^2, we can make that replacement in our formula: [tex]x= \frac{-3+/- \sqrt{i^2(11)} }{2} [/tex]. We can now pull out a single i from the i^2 and write the answer in standard form. [tex]x= \frac{-3+/-i \sqrt{11} }{2} [/tex]. [tex]x= \frac{-3+i \sqrt{11} }{2} , \frac{-3-i \sqrt{11} }{2} [/tex]. Standard form is [tex]x= \frac{-3}{2}+ \frac{i \sqrt{11} }{2}, \frac{-3}{2}- \frac{i \sqrt{11} }{2} [/tex]. There you go!
