-0.0625t^2+3t-20=0

Solution for -0.0625t^2+3t-20=0 equation:

-0.0625t^2+3t-20=0

a = -0.0625; b = 3; c = -20;
Δ = b2-4ac
Δ = 32-4·(-0.0625)·(-20)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-2}{2*-0.0625}=\frac{-5}{-0.125}
=+40
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+2}{2*-0.0625}=\frac{-1}{-0.125}
=+8
$