3n^2+28n-168=0

Solution for 3n^2+28n-168=0 equation:

3n^2+28n-168=0

a = 3; b = 28; c = -168;
Δ = b2-4ac
Δ = 282-4·3·(-168)
Δ = 2800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2800}=\sqrt{400*7}=\sqrt{400}*\sqrt{7}=20\sqrt{7}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-20\sqrt{7}}{2*3}=\frac{-28-20\sqrt{7}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+20\sqrt{7}}{2*3}=\frac{-28+20\sqrt{7}}{6}
$