4.9t^2+1.8T-5=0

Solution for 4.9t^2+1.8T-5=0 equation:

4.9t^2+1.8-5=0

We add all the numbers together, and all the variables

4.9t^2-3.2=0

a = 4.9; b = 0; c = -3.2;
Δ = b2-4ac
Δ = 02-4·4.9·(-3.2)
Δ = 62.72
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{-(0)-\sqrt{62.72}}{2*4.9}=\frac{0-\sqrt{62.72}}{9.8}
=-\frac{\sqrt{}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{62.72}}{2*4.9}=\frac{0+\sqrt{62.72}}{9.8}
=\frac{\sqrt{}}{9.8}
$