+118608 Solving Quadratic Equations by Factoring
1. 3r^2 -16r-7=5
This is a quadratice, (the highest power is 2) so there could be 0,1 or 2 real answers.
First take it all to one side by subtracting 5 from both sides.
After you do that I will look at what you have. So unless someone else answers first, the next post is yours.
+36915 3 r^2 -16r-12 = 0
eeeeek....I had an answer here...but I see Melody is coaching ya thru it.....
+118608 Anyway, EP did the first bit for you.
Now it looks pretty horrible so I would just use the quadratic formula to solve it.
\(r = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Do you know what a, b and c are?
+36915 When you have a QUADRATIC EQUATION of the form a x^2 + b x + c = 0 (or in this case a r^2 + b r + c = 0)
you can use the QUADRATIC FORMULA to find the values of x (or r in this case)
Melody showed you the FORMULA....Now do you know what a b and c are?
Put those values in the FORMULA and solve it..... OK ? 
+118608 Ok lets take a look - I'll start from the beginning.
\(3 r^2 -16r-7 = 5\\ \text{you need it to equal 0 so take five from BOTH sides}\\ 3 r^2 -16r-7-5 = 5-5\\ \color{red}{3 r^2 -16r-12 = 0}\\ \text{Now you have it in general from for a quadratic equation}\\ ar^2+br+c=0\\ \color{red}{a=3\qquad b=-16 \qquad c=-12}\\~\\ \text{Now use the quadratic formula to find the answer}\\ r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \)
Now substitute the numbers into the equation and simplify to find your answers. :)
[that was meant to be general FORM not general from ]
+128407 3r^2 - 16r - 7 = 5 subtract 5 from both sides
3r^2 - 16r - 12 = 0 we cannot take out any common factor
So....here is a procedure for factoring whenever the lead coefficient is not a 1
1) Multiply the first and last terms = 3 * -12 = -36
2) List all the possible factors of -36 and that add to - 16.....we can stop (if) we find one that adds to the middle term, -16......if we find no combination.....the polynomial is fully factored
-36 1
-18 2
Note that -18 and 2 is what we need
3) Write the original problem in two ways
3x^2 - 18x + 2x - 12 = 0 or
3x^2 + 2x - 18x - 12 = 0
The second can't be factored by grouping, but the first one can
So we have
3x^2 - 18x + 2x - 12 = 0 factor by grouping
3x ( x - 6) + 2 ( x - 6) = 0 the GCF is (x - 6)
( x - 6) ( 3x + 2) = 0 so
x - 6 = 0 or 3x + 2 = 0
x = 6 subtract 2 from both sides
3x = - 2
divide both sides by 3
x = -2/3
The solutions are in red
