1.7+12t-4.9t^2=0

Solution for 1.7+12t-4.9t^2=0 equation:

1.7+12t-4.9t^2=0

a = -4.9; b = 12; c = +1.7;
Δ = b2-4ac
Δ = 122-4·(-4.9)·1.7
Δ = 177.32
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{-(12)-\sqrt{177.32}}{2*-4.9}=\frac{-12-\sqrt{177.32}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+\sqrt{177.32}}{2*-4.9}=\frac{-12+\sqrt{177.32}}{-9.8}
$