9t(t+3)=0

Solution for 9t(t+3)=0 equation:

9t(t+3)=0

We multiply parentheses

9t^2+27t=0

a = 9; b = 27; c = 0;
Δ = b2-4ac
Δ = 272-4·9·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-27}{2*9}=\frac{-54}{18}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+27}{2*9}=\frac{0}{18}
=0
$