10.38t-4.9t^2=0

Solution for 10.38t-4.9t^2=0 equation:

10.38t-4.9t^2=0

a = -4.9; b = 10.38; c = 0;
Δ = b2-4ac
Δ = 10.382-4·(-4.9)·0
Δ = 107.7444
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{-(10.38)-\sqrt{107.7444}}{2*-4.9}=\frac{-10.38-\sqrt{107.7444}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10.38)+\sqrt{107.7444}}{2*-4.9}=\frac{-10.38+\sqrt{107.7444}}{-9.8}
$