n(n-3)=43

Solution for n(n-3)=43 equation:

n(n-3)=43

We move all terms to the left:

n(n-3)-(43)=0

We multiply parentheses

n^2-3n-43=0

a = 1; b = -3; c = -43;
Δ = b2-4ac
Δ = -32-4·1·(-43)
Δ = 181
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{181}}{2*1}=\frac{3-\sqrt{181}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{181}}{2*1}=\frac{3+\sqrt{181}}{2}
$