t^2-128t+192=0

Solution for t^2-128t+192=0 equation:

t^2-128t+192=0

a = 1; b = -128; c = +192;
Δ = b2-4ac
Δ = -1282-4·1·192
Δ = 15616
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{15616}=\sqrt{256*61}=\sqrt{256}*\sqrt{61}=16\sqrt{61}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-16\sqrt{61}}{2*1}=\frac{128-16\sqrt{61}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+16\sqrt{61}}{2*1}=\frac{128+16\sqrt{61}}{2}
$