How do I do this Question?

3x^2-10x+3

Quadratic formula

You can also factor. Me, I prefer factoring to the quad formula if it is practical. Not everything factors nicely, so the quad formula is better suited.

Anyway, 3x^{2}-10x+3=0

We have the form ax^{2}+bx+c=0

Ask yourself, "What two numbers when multiplied equal a*c and when added equal b?".

In this case, a*c=9 and b=-10.

Let's see, what could they be. How about... -9 and -1

(-9)(-1)=9 and -9+(-1)=-10

Rewrite this way:

3x^{2}\underbrace{-9x-x}_{\text{-10x}}+3=0

Group:

(3x^{2}-9x)-(x-3)=0

Factor something common out of each parentheses:

3x(x-3)-(x-3)

See? What is in both sets of parentheses is the same. That is crucial. If it is not, it will not work.

Rewrite as (3x-1)(x-3)=0

Now, it is easy to see what the zeros are.

3squareroot5 + 5squareroot6 + 3squareroot20

seeking help for the following equation
y=2/3x +1

How do I do this Question?

3x^2-10x+3
the answer is 10+13=23^+8(16)

seeking help for the following equation
y=2/3x +1

Please start your own thread for your question instead of tacking onto a thread that is not even related. You have also not stated what you are supposed to be doing with that equation. It's in slope-intercept form, but what about it?


the answer is 10+13=23^+8(16)

What?