n2-18n+45=0

Solution for n2-18n+45=0 equation:

n2-18n+45=0

We add all the numbers together, and all the variables

n^2-18n+45=0

a = 1; b = -18; c = +45;
Δ = b2-4ac
Δ = -182-4·1·45
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-12}{2*1}=\frac{6}{2}
=3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+12}{2*1}=\frac{30}{2}
=15
$