4n^2-28+196=4(n-7)

Solution for 4n^2-28+196=4(n-7) equation:

4n^2-28+196=4(n-7)

We move all terms to the left:

4n^2-28+196-(4(n-7))=0

We add all the numbers together, and all the variables

4n^2-(4(n-7))+168=0


We calculate terms in parentheses: -(4(n-7)), so:

4(n-7)

We multiply parentheses

4n-28

Back to the equation:

-(4n-28)


We get rid of parentheses

4n^2-4n+28+168=0

We add all the numbers together, and all the variables

4n^2-4n+196=0

a = 4; b = -4; c = +196;
Δ = b2-4ac
Δ = -42-4·4·196
Δ = -3120
Delta is less than zero, so there is no solution for the equation