Finding Solution to Parabolas

Examples:
1.
x^2 + x = 0
To find the solution to this equation, you need to factor out the greates common factor which in this case is x.
x(x+1) = 0
Now you need to equal each of the terms to zero and solve for x
x = 0
x+1 = 0
x = -1
Therefore the solutions to these equations are 0 and -1

2.
x^2 + 2x + 1
In this example you can't factor out any term, so you have to use factoring.
(x+1)(x+1) = 0
Like before, we need to equal these terms to 0
x+1 = 0
x = -1

3.
3x^2 + 12x + 12 = 0
At first glance it may seem very hard to factor this equation, but if you factor a 3 first it will be very simple to factor.
3(x^2+4x+4) = 0
3(x+2)(x+2) = 0
x+2=0
x=-2

4.
4x^2 + 7x - 3 = 0

In this case it seems really hard to factor and we can't factor out any term. Basically, we need to use the quadratic formula:

In the formula, a is the first constant, b the second one and c the third one (ax^2 + bx + c)Solving for x we obtain:

x = (-7 + 9.8488)/8 and x = (-7 - 9.8488)/8

x = .356, -2.106