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