n^2+46n+168=0

Solution for n^2+46n+168=0 equation:

n^2+46n+168=0

a = 1; b = 46; c = +168;
Δ = b2-4ac
Δ = 462-4·1·168
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(46)-38}{2*1}=\frac{-84}{2}
=-42
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(46)+38}{2*1}=\frac{-8}{2}
=-4
$