Question 86271
The quadratic formula is {{{x = (-b +- sqrt(b^2-4*a*c))/(2*a)}}}. In each case,it's just a matter of plugging in the appropriate values for a , b and c. Then simplify as much as possible. For #1, a = 1 , b = -1 and c = -2. So, you get :{{{x = (-(-1) +- sqrt((-1)^2-4*1*(-2)))/(2*1)}}} = {{{(1 +- sqrt(1-(-8)))/2}}} = {{{(1 +- sqrt(9))/2}}} = {{{(1 +- 3)/2}}} = {{{(1 + 3)/2}}} and {{{(1 - 3)/2}}} = {{{4/2}}} and {{{-2/2}}} = 2 and -1
For #2, you must first set it = to 0: {{{x^2}}} - 3x - 12 = 0. Then, a = 1 , b = -3 and c = -12. Now, proceed exactly as with #1 above. #s 4 and 5 likewise require that you first set them = 0. Then plug in your values for a , b and c and simplify. If you get an irrational square root, break it down and reduce the rational parts if possible. Otherwise just leave it as a square root. IF you get a 0 square root then the square root part dissappears and your answer is just x = {{{-b/(2a)}}}. If you get a negative square root, then the equation has no solution. I'll let you take it from here!