(n^2+10n-27)=3

Solution for (n^2+10n-27)=3 equation:

(n^2+10n-27)=3

We move all terms to the left:

(n^2+10n-27)-(3)=0

We get rid of parentheses

n^2+10n-27-3=0

We add all the numbers together, and all the variables

n^2+10n-30=0

a = 1; b = 10; c = -30;
Δ = b2-4ac
Δ = 102-4·1·(-30)
Δ = 220
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:

$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{55}}{2*1}=\frac{-10-2\sqrt{55}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{55}}{2*1}=\frac{-10+2\sqrt{55}}{2}
$