(n+3)(n+3)=51

Solution for (n+3)(n+3)=51 equation:

(n+3)(n+3)=51

We move all terms to the left:

(n+3)(n+3)-(51)=0

We multiply parentheses ..

(+n^2+3n+3n+9)-51=0

We get rid of parentheses

n^2+3n+3n+9-51=0

We add all the numbers together, and all the variables

n^2+6n-42=0

a = 1; b = 6; c = -42;
Δ = b2-4ac
Δ = 62-4·1·(-42)
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{51}}{2*1}=\frac{-6-2\sqrt{51}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{51}}{2*1}=\frac{-6+2\sqrt{51}}{2}
$