-16t^2+90t-120=0

Solution for -16t^2+90t-120=0 equation:

-16t^2+90t-120=0

a = -16; b = 90; c = -120;
Δ = b2-4ac
Δ = 902-4·(-16)·(-120)
Δ = 420
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:

$\sqrt{\Delta}=\sqrt{420}=\sqrt{4*105}=\sqrt{4}*\sqrt{105}=2\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-2\sqrt{105}}{2*-16}=\frac{-90-2\sqrt{105}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+2\sqrt{105}}{2*-16}=\frac{-90+2\sqrt{105}}{-32}
$