Can't wrap my head around IT. Algebra 2 HW Help :O?

A quadratic equation is one in the form:

ax^2 + bx + c

where x is the variable and c is the constant. a and b are coefficients of the variable x.

All you have to do with the problems above is get it into this form.

okay. when using the formula, you have to have the equation equal to zero. For the first one, add two to both sides, and also subtract 4x from both sides. This makes the equation become:

-3x^2 - x +2 = 0.......a= -3 ; b= -1 ; c=2

The next one you need to add 6x^2 to both sides and subtract 1. This makes the equation become:

13x^2 + x -1 = 0 ........ a=13 ; b=1 ; c= -1

Plug those numbers into the quadratic formula as you would normally do and you should be sweeeet. Hope this helps!

Move the common terms to one side such that one side of the equation is zero. First step.

Problem 1)

-3x^2+3x=4x-2

-3x^2-x=-2

-3x^2-x+2=0

a=-3, b=-1, c=2

->Alternatively, you can factor this out to become:

-(3x^2+x-2)

-(3x-2)(x+1)=0

so the zeros of the function are (2/3) and -1.

Problem 2)

7x^2+x=1-6x^2

13x^2+x=1

13x^2+x-1=0

a=13, b=1, c=-1

->This cannot be factored. You will either use the quadratic function (easier) or complete the square.

(-1±7)/2 is the answer.

The quadratic formula works when you are solving a quadratic polynomial (ie. the highest exponent on x is 2), and one side is equal to zero. So... move the stuff on the right side over to the left side (or the other way around if you really want, it doesn't matter), and from there just use the quadratic formula that you already know how to use.

all you need to do is get the form of ax^2 + bx + c = 0

So when you move all the variables into one side of the equation,

you get 0 = 3x^2 + x - 2 which comes out (3x-2)(x+1)=0

For the 2nd qn you get 0 = 13x^2 + x -1. Use (-b+(b^2-4ac)^1/2)/2a and (-b-(b^2-4ac)^1/2)/2a to get two different values of x.

it goes to the form:

-3x^2-x+2=0

solve with formula