-10t^2+80t-150=0

Solution for -10t^2+80t-150=0 equation:

-10t^2+80t-150=0

a = -10; b = 80; c = -150;
Δ = b2-4ac
Δ = 802-4·(-10)·(-150)
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{-(80)-20}{2*-10}=\frac{-100}{-20}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+20}{2*-10}=\frac{-60}{-20}
=+3
$