(4n)^2-52n+168=0

Solution for (4n)^2-52n+168=0 equation:

(4n)^2-52n+168=0

a = 4; b = -52; c = +168;
Δ = b2-4ac
Δ = -522-4·4·168
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-4}{2*4}=\frac{48}{8}
=6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+4}{2*4}=\frac{56}{8}
=7
$