What is the product of 4n^+3=7n?

So if you solve for n
4n^2+3=7n
make into trinomial
subtract 7n from both sides
4n^2-7n+3=0
if we can factor this, then we can asume that the factors are equal to zero becase if
xy=0 then assume x and/or y=0 so

to factor an equation in ax^2+bx+c where a is greater than 1 then
b=t+z
a times c=t times z
to factor 4n^2-7n+3 you do
4 times 3=12
what 2 numbers multiply to get 12 and add to get -7
the numbers are -3 and -4 so
split up the -7
4n^2-4n-3n+3
group
(4n^2-4n)+(-3n+3)
undistribute
(4n)(n-1)+(-3)(n-1)
reverse distributive property
ab+ac=a(b+c)
(4n-3)(n-1)=0
set each to zero
4n-3=0
add 3 to both sides
4n=3
divide both sides by 4
n=3/4

n-1=0
add 1 to both sides
n=1

n=3/4 or 1

It would be the second one because it passes the vertical line test