(N+2)x2=20

Solution for (N+2)x2=20 equation:

(+2)N2=20

We move all terms to the left:

(+2)N2-(20)=0

We add all the numbers together, and all the variables

2N2-20=0

We add all the numbers together, and all the variables

2N^2-20=0

a = 2; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·2·(-20)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*2}=\frac{0-4\sqrt{10}}{4}
=-\frac{4\sqrt{10}}{4}
=-\sqrt{10}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*2}=\frac{0+4\sqrt{10}}{4}
=\frac{4\sqrt{10}}{4}
=\sqrt{10}
$