n2-3n+8,n=4

Solution for n2-3n+8,n=4 equation:

n2-3n+8.n=4

We move all terms to the left:

n2-3n+8.n-(4)=0

We add all the numbers together, and all the variables

n^2+5n-4=0

a = 1; b = 5; c = -4;
Δ = b2-4ac
Δ = 52-4·1·(-4)
Δ = 41
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{-(5)-\sqrt{41}}{2*1}=\frac{-5-\sqrt{41}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{41}}{2*1}=\frac{-5+\sqrt{41}}{2}
$