-5t^2-2t+36=0

Solution for -5t^2-2t+36=0 equation:

-5t^2-2t+36=0

a = -5; b = -2; c = +36;
Δ = b2-4ac
Δ = -22-4·(-5)·36
Δ = 724
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:

$\sqrt{\Delta}=\sqrt{724}=\sqrt{4*181}=\sqrt{4}*\sqrt{181}=2\sqrt{181}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{181}}{2*-5}=\frac{2-2\sqrt{181}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{181}}{2*-5}=\frac{2+2\sqrt{181}}{-10}
$