4-t^2+80t-100=0

Solution for 4-t^2+80t-100=0 equation:

4-t^2+80t-100=0

We add all the numbers together, and all the variables

-1t^2+80t-96=0

a = -1; b = 80; c = -96;
Δ = b2-4ac
Δ = 802-4·(-1)·(-96)
Δ = 6016
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:

$\sqrt{\Delta}=\sqrt{6016}=\sqrt{64*94}=\sqrt{64}*\sqrt{94}=8\sqrt{94}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-8\sqrt{94}}{2*-1}=\frac{-80-8\sqrt{94}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+8\sqrt{94}}{2*-1}=\frac{-80+8\sqrt{94}}{-2}
$