0=-3t2+15t

Solution for 0=-3t2+15t equation:

0=-3t^2+15t

We move all terms to the left:

0-(-3t^2+15t)=0

We add all the numbers together, and all the variables

-(-3t^2+15t)=0

We get rid of 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{0}{6}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*3}=\frac{30}{6}
=5
$