-16t^2+270t-360=0

Solution for -16t^2+270t-360=0 equation:

-16t^2+270t-360=0

a = -16; b = 270; c = -360;
Δ = b2-4ac
Δ = 2702-4·(-16)·(-360)
Δ = 49860
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:

$\sqrt{\Delta}=\sqrt{49860}=\sqrt{36*1385}=\sqrt{36}*\sqrt{1385}=6\sqrt{1385}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(270)-6\sqrt{1385}}{2*-16}=\frac{-270-6\sqrt{1385}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(270)+6\sqrt{1385}}{2*-16}=\frac{-270+6\sqrt{1385}}{-32}
$