Question

solve using the quadratic formula. I'm trying to graduate, so guys or girls please help me.
3x^2-8x+2=0

Answer 1

The solution for the general equation,
\[ax^2+bx+c=0\] is \[x=\frac{b\pm\sqrt{b^2-4ac}}{2a}\] You can do it from this point.

Answer 2

I'm not understanding this formula.

Answer 3

Compare the general equation with your equation. You will see that,
\[a=3\\ b=8 \\ c=2\] Then substitute in the solution,
\[x=\frac{-8\pm\sqrt{8^2-4\cdot3\cdot 2}}{2\cdot 3}\] You will have two solutions, one for each sign of the square root,
\[x=\frac{-8-\sqrt{8^2-4\cdot3\cdot 2}}{2\cdot 3}\\\]

Answer 4

And
\[x=\frac{-8+\sqrt{8^2-4\cdot3\cdot 2}}{2\cdot 3}\]

Answer 5

?

Answer 6

In the formula it should be [-b. +- sqrt(b^2-4ac)]/2a first - missing
Example worked properly

Answer 7

right, but how do i work these types of problems?

Answer 8

re-write them so that they fit the format
ax^2 + bx + c = 0

then use formula for solving quadratic equations, as given above

note that there will be two solutions because of the +- sign in the formula.

For a real problem, one of these solutions will make sense, the other will not.

If the sqrt has a negative b^2 - 4 a c, then you will have tow "imaginary" solutions,
with format x = something +- i somethingelse

Answer 9

I'm sorry but I'm very very lost right now.

Answer 10

What is the question you have about what has been presented?

Answer 11

all it says is to put the Approximate answers to the nearest tenth.

Answer 12

Use the formula, get an answer, round it up or down, depending on whether
the decimal part of the answer is greater or less than 0.A5:
<0.A5 becomes 0.A rounded down
>= 0.A5 becomes 0.(A+1) rounded up

e.g., 2.15 = 2.2
2.14 = 2.1

Answer 13

?

Answer 14

x = 2.39, 0.28

Answer 15

oh........ok