0=-5t^2+21t

Solution for 0=-5t^2+21t equation:

0=-5t^2+21t

We move all terms to the left:

0-(-5t^2+21t)=0

We add all the numbers together, and all the variables

-(-5t^2+21t)=0

We get rid of parentheses

5t^2-21t=0

a = 5; b = -21; c = 0;
Δ = b2-4ac
Δ = -212-4·5·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21}{2*5}=\frac{0}{10}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21}{2*5}=\frac{42}{10}
=4+1/5
$