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In order to solve a quadratic equation by factoring, follow these steps:

1. Put the equation in the standard form (ax2+bx+c=0)
2. Factor the equation (find two numbers that will not only multiply to equal the constant term "c", but also add up to equal "b", the coefficient on the x-term).
3. Set each of the two binomial expressions equal to cero.
4. Solve each of the equations.

Solve by factoring x2+2x = 15
1. Put the equation in the standard form: x2+2x-15=0
2. We require to numbers that multiply together to give -15 and add together to give -2.
x2+2x-15=(x-3)(x+5).

3. Set each of the two binomial expressions equal to cero.
x2+2x-15 = 0
(x-3)(x+5) =0
4. Solve each equation:
x-3=0, x=3
x+5=0, x=-5

When the leading coefficient (the number on the x2 term) is not 1, the first step in factoring will be to multiply "a" and "c"; then we'll need to find factors of the product "ac" that add up to "b".

Solve by factoring 2x2+4x-6 = 0
We need to find factors of 12 (ac=2·(-6)=-12) that add up to +4.
We will use the pair "-2 and 6".
Draw a two-by-two grid, putting the first term in the upper left-hand corner and the last term in the lower right-hand corner:

 2x2 -6

Take the factors –2 and 6 and put them, complete with their signs and variables, in the diagonal corners:
 2x2 -2x 6x -6
Factor the rows and columns:

 2x -2 2x 2x2 -2x 6 6x -6

Then, 2x2+4x-6 = (2x-2)(2x+6)
We will find the solutions of the equations by solving each equation:
2x-2=0, x=1
2x+6=0, x=-3

Sometimes you can not find integer factors that work, then this quadratic is said to be "unfactorable over the integers" or "prime". On these cases, you must try to solve the equation using another method (the quadratic formula,...)

Solve $\fs1-7x^2+2x-3=0$

No solution
One solution x=
Two solutions x1= x2=