t^2-11.25t-19.83=0

Solution for t^2-11.25t-19.83=0 equation:

t^2-11.25t-19.83=0

a = 1; b = -11.25; c = -19.83;
Δ = b2-4ac
Δ = -11.252-4·1·(-19.83)
Δ = 205.8825
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{-(-11.25)-\sqrt{205.8825}}{2*1}=\frac{11.25-\sqrt{205.8825}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11.25)+\sqrt{205.8825}}{2*1}=\frac{11.25+\sqrt{205.8825}}{2}
$