n(n-3)=340

Solution for n(n-3)=340 equation:

n(n-3)=340

We move all terms to the left:

n(n-3)-(340)=0

We multiply parentheses

n^2-3n-340=0

a = 1; b = -3; c = -340;
Δ = b2-4ac
Δ = -32-4·1·(-340)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-37}{2*1}=\frac{-34}{2}
=-17
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+37}{2*1}=\frac{40}{2}
=20
$