192=t(128-t)

Solution for 192=t(128-t) equation:

192=t(128-t)

We move all terms to the left:

192-(t(128-t))=0

We add all the numbers together, and all the variables

-(t(-1t+128))+192=0


We calculate terms in parentheses: -(t(-1t+128)), so:

t(-1t+128)

We multiply parentheses

-1t^2+128t

Back to the equation:

-(-1t^2+128t)


We get rid of parentheses

1t^2-128t+192=0

We add all the numbers together, and all the variables

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}
$