3(t^2-6t+5)=0

Solution for 3(t^2-6t+5)=0 equation:

3(t^2-6t+5)=0

We multiply parentheses

3t^2-18t+15=0

a = 3; b = -18; c = +15;
Δ = b2-4ac
Δ = -182-4·3·15
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-12}{2*3}=\frac{6}{6}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+12}{2*3}=\frac{30}{6}
=5
$