Question

how you solve 3v^2+6=11v

Answer 1

Type your equation into Wolfram Alpha.

Answer 2

haha
use the quadratic equation after you have moved the 11v to the left side.

Answer 3

Make the equation equal to zero.
3v^2+6=11v
3v^2-11v+6=0

Use the quadratic formula:
v= -b+/-sqrt(b^2-4ac)/2a
a=3
b=-11
c=6

v=2/3
v=3

Answer 4

This one is factorable, also.

Answer 5

i need help factoring it

Answer 6

how you do solve 15u=4u^2-4

Answer 7

First set it equal to 0\[3v^2-11v+6=0\]It will be factorable if factors of the product of the lead coefficient, 3 and the constant, 6 will add up to -11, and there is such a pair:
(-9)(-2)=18
-9-2=-11
Now you can use these factors to rewrite the -11v as -9v-2v:\[=3v^2-9v-2v+6\]Now you can factor by grouping (which means factor out the GCF of the first two terms and the second two terms, if possible):\[=3v(v-3)-2(v-3)\]Notice now that there are two terms each with the GCF v-3. So factor the binomial out which leaves the 3v and the -2\[=(v-3)(3v-2)\]This is the factored form (you can check it my FOIL-ing).
(continued)

Answer 8

Remember that this was an equation set equal to zero (sorry i got a little sloppy). So we should have\[(v-3)(3v-2)=0\]Each of these factors may be set equal to zero to solve\[v-3=0 \ or \ 3v-2=0\]\[v=3 \ or \ v=2/3\]these are the answers

Answer 9

how you do solve 15u=4u^2-4
Set equal to zero\[4u^2-15u-4=0\]rewrite -15 using factors of (4)(-4)=-16 which would be -16 and +1\[4u^2-16u+u-4=0\]Factor by grouping\[4u(u-4)+(u-4)=0\]Factor out the GCF binomial\[(u-4)(4u+1)=0\]set each factor equal to zero and solve for u\[u-4=0 \ or \ 4u+1=0\]\[u=4 \ or\ u=-1/4\]