3t(t+5)=0

Solution for 3t(t+5)=0 equation:

3t(t+5)=0

We multiply parentheses

3t^2+15t=0

a = 3; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·3·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*3}=\frac{-30}{6}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*3}=\frac{0}{6}
=0
$