-4.9t^2+16t-3.7=0

Solution for -4.9t^2+16t-3.7=0 equation:

-4.9t^2+16t-3.7=0

a = -4.9; b = 16; c = -3.7;
Δ = b2-4ac
Δ = 162-4·(-4.9)·(-3.7)
Δ = 183.48
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{-(16)-\sqrt{183.48}}{2*-4.9}=\frac{-16-\sqrt{183.48}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+\sqrt{183.48}}{2*-4.9}=\frac{-16+\sqrt{183.48}}{-9.8}
$