Are there any shortcuts to quadratic functions?
961776
That is not the quadratic formula it is -b plus or minus, b squared - 4(ac) all over 2a, you would best to just look up the quadratic formula to understand it, you can also use factorization, or completeing the square, but must have a coefficient of one.  For larger pol nomials there are other ways, but just best to google it or yahoo it considering the site i am on.
538847
For any quadratic of the form ax^2 + bx + c = 0, the roots are\n\nx = (-b +/- sqrt(b^2-4ac)) / 2a\n\nSqrt here is shorthand for "the square root of".
374428
The most strait forward way is to use the quadratic formula. You just do simple adding, powering, rooting, and you get the answer. It works the 100% of the time. There is also the filling-in-the-square, but that's too much work compaired to the quadratic formula.
464622
Use any recursive algorithm.\n\nFor instance:\n\nf(x) = f(x) + 1
913334
x^2 + 2x + 7 Remainder 10\n\ni can add the steps if you don't think i did it by hand
952526
x^2-5x-2=0\n\nuse the quadratic formula\n\n[5  +/- sqrt(25-4*1*-2)]/2\n\n[5  +/-   sqrt(25+ 8)]/2\n\nI distributed the 1/2...easier to type\n\n5/2  +/-sqrt33/2
152856
There are lots of methods for approximating roots by hand. \n\nFor example, here is a more general technique, called the Newton-Raphson method. \n\nLet F(X)=(X^2)-2\n\nYou want to find the place R where F(R)=0.\n\nSo to find the roots of F(X), iterate. \n\n\nXn+1=Xn-F(Xn)/F'(Xn)\n\nwhere F'(X)=2X\n\nPick a starting value for X\n\nX0=1, say, and then compute X1, then X2, and so on. \nIt should converge pretty rapidly. \n\nAnd it works for other fractional powers too !
913819
The "Hindu Solution" to the quadratic equation is the same method that we now call "completing the square."  This is a way to convert terms in the equation to the form of the square of a different polynomial, enabling it to be solved algebraically.  When applied to a quadratic equation in the general form, this method yields the quadratic formula.
175327
Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
676950
A few that come to mind at Euler, Maxima and Octave. Unfortunately, I don't think any of the free packages come close to Mathematica or Matlab, but maybe one of them will still meet your needs.
87719
x^2 + 9x + 10x + 90\n\n(x + 9)(x + 10)
858418
Click ON\nClick PRGM\nSelect NEW tab and hit ENTER\n\nTo select DISP and PROMPT:\n Hit PRGM when creating or editing a program,\n Select Tab I/O and you will see you selection\n\nTo get the arrow, click STO-> just to the left of 1\n\n\nThis is the Program: (or the one I use)\n\n:Disp "AX^2+BX+C"\n\n:Prompt A\n:Prompt B\n:Prompt C\n:(-B+*square root*(B^2-4AC))/(2A)->D\n:(-B-*square root*(B^2-4AC))/(2A)->E\n:Disp D\n:Disp E\n\nTo run program, go to the main screen, click PRGM, select your program, hit ENTER, input values for A, B, and C, hit ENTER and you result will show your two possible answers.
1150325
The quadratic formula is:  [-b+/-((b^2-4ac))/(2a)] where a is the number befor x^2, b is the number before x and c is the integer.    Using this formula we get:\n\n1. x = [1+/- (1-(4)(6)(-12))^1/2/((2)(6))]\n    x = [1+/- 17]/(12)\n    x = 18/12 or -16/12\n    x = 1.5, x = -4/3\n\n2. x =  [6+/- (36-(4)(1)(13))^1/2/((2)(1))]\n    x = [6 +/- (16)^(1/2) i}(2) {since there is a negative sign inside the sq root we replace it by and positive sq root and imaginary number}.\n    x = [6 +/- 4i]/(2)\n   x = 3 + 2i, x = 3-2i
817524
For a quadratic equation, the maximum or minimum will be the vertex. Since the first term of the quadratic is positive, the parabola goes up, so you're looking for the minimum.\n\nThere are 2 methods:\nCalculus (using derivatives):\nf(x) = x2 - 2x - 4\nf'(x) = 2x - 2\nthe maximum/minimum will be where f'(x) = 0.\n2x - 2 = 0\nx = 1\ny = 1^2 - 2*1 - 4 = -5\n\nAlgebra:\nThe vertex can be found using the formula:\n(-b/2a , -(b^2 - 4ac)/4a)\n(-(-2)/2, -(4 - (4*1*-4))/4)\n(1,-5)\n\nCheck: x = 1, y = 1^2 - 2*1 - 4 = -5
917477
where else?\n\nhttp://www.factoring-polynomials.com/
82007
Just type in quadratic equation and maybe roots in a serch engine. You will have your pick of dozens of sites.
922434
There is only one step. Take the quadratic formula and plug in the numbers.\n\nIf ax^2 + bx + c = 0,\nx = (-b +/- sqrt(b^2-4ac))/2a\n\nFor the equation you gave, a=1, b=19, c=88. Plug those numbers in and see what you get.\n\nIf you're confused by the +/-, it's shorthand for writing two things in one. There are usually two values of x that work, one with + and one with -.