140=80t-5*(t^2)

Solution for 140=80t-5*(t^2) equation:

140=80t-5(t^2)

We move all terms to the left:

140-(80t-5(t^2))=0

We get rid of parentheses

5t^2-80t+140=0

a = 5; b = -80; c = +140;
Δ = b2-4ac
Δ = -802-4·5·140
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-60}{2*5}=\frac{20}{10}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+60}{2*5}=\frac{140}{10}
=14
$