-16t^2+90t+4.5=0

Solution for -16t^2+90t+4.5=0 equation:

-16t^2+90t+4.5=0

a = -16; b = 90; c = +4.5;
Δ = b2-4ac
Δ = 902-4·(-16)·4.5
Δ = 8388
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:

$\sqrt{\Delta}=\sqrt{8388}=\sqrt{36*233}=\sqrt{36}*\sqrt{233}=6\sqrt{233}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-6\sqrt{233}}{2*-16}=\frac{-90-6\sqrt{233}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+6\sqrt{233}}{2*-16}=\frac{-90+6\sqrt{233}}{-32}
$