Quadratic Equations, believe it or not, are not that hard. Let's take a look at how to solve one:

Equation: 4x^{2}+4x-3=2

First off, we want to make the 2 in the equation a zero*, so we subtract 2 from both sides:

2-2=0

-3-2=-5

*You don't need to do this if the right number is already zero.

And we'll have the equation we need: 4x^{2}+4x-5=

Now, we assign the numbers to variables of their own using this equation: Ax^{2}+Bx+C. In this case:

A = 4

B = 4

C = -5

To solve for our original variable, x, we need to use the quadratic formula:

x=(-B±√(B^{2}-4AC))/2A

What this means is that there are going to be 2 solutions. Let's solve for X by plugging in A, B, and C:

x=(-4±√(4^{2}-4(4)(-5)))/2(4)

I always start by solving what's in the radical.

4^{2}-4(4)(-5)

We start with Exponents since there are no multi-term expressions in parentheses.

4^{2}=16

Then Multiplication:

-4*4=-16

-16*-5=80

We now have 16+80 which equals 96. We now have that.

x=-4±√(96)/2(4)

Let's multiply the bottom side of the fraction:

2*4=8

x=-4±√(96)/8

For all intents and purposes, I'm converting √(96) to a decimal.

√96 ≈ 9.798

x=-4±9.798/8

We now turn this into 2 equations:

x=-4+9.798/8

x=-4-9.798/8

Then combine terms on the numerator:

x=5.798/8

x=-13.798/8

Then divide.

x=0.72475

x=-1.72475.

So x indeed does have 2 solutions.

x=0.72475,-1.72475

I hope you enjoyed this eyesore as much as I did