16x^2-192x+320=0

Solution for 16x^2-192x+320=0 equation:

16x^2-192x+320=0

a = 16; b = -192; c = +320;
Δ = b2-4ac
Δ = -1922-4·16·320
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-192)-128}{2*16}=\frac{64}{32}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-192)+128}{2*16}=\frac{320}{32}
=10
$