320-80t-5t2=0

Solution for 320-80t-5t2=0 equation:

320-80t-5t^2=0

a = -5; b = -80; c = +320;
Δ = b2-4ac
Δ = -802-4·(-5)·320
Δ = 12800
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:

$\sqrt{\Delta}=\sqrt{12800}=\sqrt{6400*2}=\sqrt{6400}*\sqrt{2}=80\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-80\sqrt{2}}{2*-5}=\frac{80-80\sqrt{2}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+80\sqrt{2}}{2*-5}=\frac{80+80\sqrt{2}}{-10}
$