0=t(10-4.905t)

Solution for 0=t(10-4.905t) equation:

0=t(10-4.905t)

We move all terms to the left:

0-(t(10-4.905t))=0

We add all the numbers together, and all the variables

-(t(-4.905t+10))+0=0

We add all the numbers together, and all the variables

-(t(-4.905t+10))=0


We calculate terms in parentheses: -(t(-4.905t+10)), so:

t(-4.905t+10)

We multiply parentheses

-4t^2+10t

Back to the equation:

-(-4t^2+10t)


We get rid of parentheses

4t^2-10t=0

a = 4; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·4·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2*4}=\frac{0}{8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*4}=\frac{20}{8}
=2+1/2
$