0(t)=-4.9t^2+16t-8

Solution for 0(t)=-4.9t^2+16t-8 equation:

0(t)=-4.9t^2+16t-8

We move all terms to the left:

0(t)-(-4.9t^2+16t-8)=0

We add all the numbers together, and all the variables

-(-4.9t^2+16t-8)+t=0

We get rid of parentheses

4.9t^2-16t+t+8=0

We add all the numbers together, and all the variables

4.9t^2-15t+8=0

a = 4.9; b = -15; c = +8;
Δ = b2-4ac
Δ = -152-4·4.9·8
Δ = 68.2
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{68.2}}{2*4.9}=\frac{15-\sqrt{68.2}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{68.2}}{2*4.9}=\frac{15+\sqrt{68.2}}{9.8}
$