-4.9t^2+27t+35=0

Solution for -4.9t^2+27t+35=0 equation:

-4.9t^2+27t+35=0

a = -4.9; b = 27; c = +35;
Δ = b2-4ac
Δ = 272-4·(-4.9)·35
Δ = 1415
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{-(27)-\sqrt{1415}}{2*-4.9}=\frac{-27-\sqrt{1415}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{1415}}{2*-4.9}=\frac{-27+\sqrt{1415}}{-9.8}
$