8n(28-n)=7(n-4)

Solution for 8n(28-n)=7(n-4) equation:

8n(28-n)=7(n-4)

We move all terms to the left:

8n(28-n)-(7(n-4))=0

We add all the numbers together, and all the variables

8n(-1n+28)-(7(n-4))=0

We multiply parentheses

-8n^2+224n-(7(n-4))=0


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

7(n-4)

We multiply parentheses

7n-28

Back to the equation:

-(7n-28)


We get rid of parentheses

-8n^2+224n-7n+28=0

We add all the numbers together, and all the variables

-8n^2+217n+28=0

a = -8; b = 217; c = +28;
Δ = b2-4ac
Δ = 2172-4·(-8)·28
Δ = 47985
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(217)-\sqrt{47985}}{2*-8}=\frac{-217-\sqrt{47985}}{-16}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(217)+\sqrt{47985}}{2*-8}=\frac{-217+\sqrt{47985}}{-16}
$