n2+3=8,n=6

Solution for n2+3=8,n=6 equation:

n2+3=8.n=6

We move all terms to the left:

n2+3-(8.n)=0

We add all the numbers together, and all the variables

n2-(+8.n)+3=0

We add all the numbers together, and all the variables

n^2-(+8.n)+3=0

We get rid of parentheses

n^2-8.n+3=0

We add all the numbers together, and all the variables

n^2-8n+3=0

a = 1; b = -8; c = +3;
Δ = b2-4ac
Δ = -82-4·1·3
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{13}}{2*1}=\frac{8-2\sqrt{13}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{13}}{2*1}=\frac{8+2\sqrt{13}}{2}
$