N+5=(16/4)x2

Solution for N+5=(16/4)x2 equation:

+5=(16/4)N2

We move all terms to the left:

+5-((16/4)N2)=0


Domain of the equation: 4)N2)!=0

N!=0/1

N!=0

N∈R


We add all the numbers together, and all the variables

-(4N2)+5=0

We add all the numbers together, and all the variables

-4N^2+5=0

a = -4; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·(-4)·5
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*-4}=\frac{0-4\sqrt{5}}{-8}
=-\frac{4\sqrt{5}}{-8}
=-\frac{\sqrt{5}}{-2}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-4}=\frac{0+4\sqrt{5}}{-8}
=\frac{4\sqrt{5}}{-8}
=\frac{\sqrt{5}}{-2}
$