N=(4.2n)(2n-7)

Solution for N=(4.2n)(2n-7) equation:

=(4.2N)(2N-7)

We move all terms to the left:

-((4.2N)(2N-7))=0

We add all the numbers together, and all the variables

-((+4.2N)(2N-7))=0

We multiply parentheses ..

-((+8N^2-28N))=0


We calculate terms in parentheses: -((+8N^2-28N)), so:

(+8N^2-28N)

We get rid of parentheses

8N^2-28N

Back to the equation:

-(8N^2-28N)


We get rid of parentheses

-8N^2+28N=0

a = -8; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·(-8)·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*-8}=\frac{-56}{-16}
=3+1/2
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*-8}=\frac{0}{-16}
=0
$