n^2+3n=3

Solution for n^2+3n=3 equation:

n^2+3n=3

We move all terms to the left:

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

a = 1; b = 3; c = -3;
Δ = b2-4ac
Δ = 32-4·1·(-3)
Δ = 21
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{-(3)-\sqrt{21}}{2*1}=\frac{-3-\sqrt{21}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{21}}{2*1}=\frac{-3+\sqrt{21}}{2}
$