4.9t^2+1.8t-52=0

Solution for 4.9t^2+1.8t-52=0 equation:

4.9t^2+1.8t-52=0

a = 4.9; b = 1.8; c = -52;
Δ = b2-4ac
Δ = 1.82-4·4.9·(-52)
Δ = 1022.44
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{-(1.8)-\sqrt{1022.44}}{2*4.9}=\frac{-1.8-\sqrt{1022.44}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.8)+\sqrt{1022.44}}{2*4.9}=\frac{-1.8+\sqrt{1022.44}}{9.8}
$