t^2-80t+1400=0

Solution for t^2-80t+1400=0 equation:

t^2-80t+1400=0

a = 1; b = -80; c = +1400;
Δ = b2-4ac
Δ = -802-4·1·1400
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-20\sqrt{2}}{2*1}=\frac{80-20\sqrt{2}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+20\sqrt{2}}{2*1}=\frac{80+20\sqrt{2}}{2}
$