OpenStudy (anonymous):

i just need the answer choice. I know how to get it. just not sure. Solve. 3x2 = -15x + 21 answer choices: the quantity of negative 5 plus or minus i square root 3 all over 2 the quantity of 5 plus or minus square root 53 all over 2 the quantity of 5 plus or minus 3 square root 3 all over 2 the quantity of negative 5 plus or minus square root 53 all over 2

OpenStudy (mertsj):

1. Get it equal to 0 2. Divide both sides by 3 3. Identify a, b, and c 4. Plug into the quadratic ormula.

OpenStudy (anonymous):

which one is it though...?

OpenStudy (whpalmer4):

Follow those steps. Do you need help doing so?

OpenStudy (whpalmer4):

Those steps are exactly how we would find the answer to give to you, so why don't you do it and get the practice?

OpenStudy (radar):

When you completed step 4, that would be your answer.

OpenStudy (radar):

Show your work for step 4, then we can analyze your problem.

OpenStudy (anonymous):

okay, question, do I put the right side equal to zero or he left?

OpenStudy (mertsj):

The right.

OpenStudy (anonymous):

okaaaay..x=7/5?

OpenStudy (mertsj):

Could you show how you got that?

OpenStudy (anonymous):

-15x+21=0 15x=-21 x=7/5

OpenStudy (whpalmer4):

You misunderstood the directions: you need to rearrange the equation so that everything is the left, and 0 on the right. You still have all three terms!

OpenStudy (anonymous):

OHHH! thanks!

OpenStudy (whpalmer4):

:-)

OpenStudy (anonymous):

so its going to be 3x^2+15x-21=0

OpenStudy (whpalmer4):

yes

OpenStudy (anonymous):

then what do I do?

OpenStudy (mertsj):

Step 2

OpenStudy (whpalmer4):

Now you have your quadratic in the form desired: ax^2+bx+c = 0. plug values into quadratic formula \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

OpenStudy (mertsj):

Make your life simpler and divide both sides by 3 first.

OpenStudy (anonymous):

do I not simplify them because they are divisible by 3..?

OpenStudy (whpalmer4):

you can if you want. makes no difference to the answer.

OpenStudy (anonymous):

so I got s(x^2+5x-7)=0

OpenStudy (anonymous):

supposed to be a 3 in the begining

OpenStudy (whpalmer4):

yeah. so now your values of a,b,c are?

OpenStudy (anonymous):

a=3 b=5 c=-7?

OpenStudy (anonymous):

a=1

OpenStudy (whpalmer4):

you divide by 3 to get rid of that leading 3... right, a = 1, b = 5, c = -7

OpenStudy (anonymous):

thanks so much! I got it!

OpenStudy (whpalmer4):

which choice is the correct answer?

OpenStudy (anonymous):

D

OpenStudy (whpalmer4):

very good!

OpenStudy (anonymous):

:)!!!!!!!!

OpenStudy (whpalmer4):

go forth and solve many quadratics :-)

OpenStudy (whpalmer4):

enjoy the fact that you get that nice, simple formula to use. It all goes out the window when you start doing x^3...