4t-4.9t^2+.45=0

Solution for 4t-4.9t^2+.45=0 equation:

4t-4.9t^2+.45=0

We add all the numbers together, and all the variables

-4.9t^2+4t+0.45=0

a = -4.9; b = 4; c = +0.45;
Δ = b2-4ac
Δ = 42-4·(-4.9)·0.45
Δ = 24.82
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{-(4)-\sqrt{24.82}}{2*-4.9}=\frac{-4-\sqrt{24.82}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{24.82}}{2*-4.9}=\frac{-4+\sqrt{24.82}}{-9.8}
$