t^2=9T=20=0

Solution for t^2=9T=20=0 equation:

t^2=9=20=0

We move all terms to the left:

t^2-(9)=0

a = 1; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·1·(-9)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*1}=\frac{-6}{2}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*1}=\frac{6}{2}
=3
$