16n^2-160n+180=0

Solution for 16n^2-160n+180=0 equation:

16n^2-160n+180=0

a = 16; b = -160; c = +180;
Δ = b2-4ac
Δ = -1602-4·16·180
Δ = 14080
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:

$\sqrt{\Delta}=\sqrt{14080}=\sqrt{256*55}=\sqrt{256}*\sqrt{55}=16\sqrt{55}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-16\sqrt{55}}{2*16}=\frac{160-16\sqrt{55}}{32}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+16\sqrt{55}}{2*16}=\frac{160+16\sqrt{55}}{32}
$