-0.8t^2+36t+2=0

Solution for -0.8t^2+36t+2=0 equation:

-0.8t^2+36t+2=0

a = -0.8; b = 36; c = +2;
Δ = b2-4ac
Δ = 362-4·(-0.8)·2
Δ = 1302.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-\sqrt{1302.4}}{2*-0.8}=\frac{-36-\sqrt{1302.4}}{-1.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+\sqrt{1302.4}}{2*-0.8}=\frac{-36+\sqrt{1302.4}}{-1.6}
$