-5t^2+40t-60=0

Solution for -5t^2+40t-60=0 equation:

-5t^2+40t-60=0

a = -5; b = 40; c = -60;
Δ = b2-4ac
Δ = 402-4·(-5)·(-60)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20}{2*-5}=\frac{-60}{-10}
=+6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20}{2*-5}=\frac{-20}{-10}
=+2
$