-1.8t^2+20t+90=0

Solution for -1.8t^2+20t+90=0 equation:

-1.8t^2+20t+90=0

a = -1.8; b = 20; c = +90;
Δ = b2-4ac
Δ = 202-4·(-1.8)·90
Δ = 1048
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:

$\sqrt{\Delta}=\sqrt{1048}=\sqrt{4*262}=\sqrt{4}*\sqrt{262}=2\sqrt{262}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{262}}{2*-1.8}=\frac{-20-2\sqrt{262}}{-3.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{262}}{2*-1.8}=\frac{-20+2\sqrt{262}}{-3.6}
$