K(t)=1.6+5t-4.9t^2

Solution for K(t)=1.6+5t-4.9t^2 equation:

(K)=1.6+5K-4.9K^2

We move all terms to the left:

(K)-(1.6+5K-4.9K^2)=0

We get rid of parentheses

4.9K^2-5K+K-1.6=0

We add all the numbers together, and all the variables

4.9K^2-4K-1.6=0

a = 4.9; b = -4; c = -1.6;
Δ = b2-4ac
Δ = -42-4·4.9·(-1.6)
Δ = 47.36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-\sqrt{47.36}}{2*4.9}=\frac{4-\sqrt{47.36}}{9.8}
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{47.36}}{2*4.9}=\frac{4+\sqrt{47.36}}{9.8}
$