6n^2+48n+42=0

Solution for 6n^2+48n+42=0 equation:

6n^2+48n+42=0

a = 6; b = 48; c = +42;
Δ = b2-4ac
Δ = 482-4·6·42
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-36}{2*6}=\frac{-84}{12}
=-7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+36}{2*6}=\frac{-12}{12}
=-1
$