3t(t-6)=0

Solution for 3t(t-6)=0 equation:

3t(t-6)=0

We multiply parentheses

3t^2-18t=0

a = 3; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·3·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*3}=\frac{0}{6}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*3}=\frac{36}{6}
=6
$