1+2.7t-4.9t^2=0

Solution for 1+2.7t-4.9t^2=0 equation:

1+2.7t-4.9t^2=0

a = -4.9; b = 2.7; c = +1;
Δ = b2-4ac
Δ = 2.72-4·(-4.9)·1
Δ = 26.89
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{-(2.7)-\sqrt{26.89}}{2*-4.9}=\frac{-2.7-\sqrt{26.89}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.7)+\sqrt{26.89}}{2*-4.9}=\frac{-2.7+\sqrt{26.89}}{-9.8}
$