-4.9t^2+11t+34=0

Solution for -4.9t^2+11t+34=0 equation:

-4.9t^2+11t+34=0

a = -4.9; b = 11; c = +34;
Δ = b2-4ac
Δ = 112-4·(-4.9)·34
Δ = 787.4
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{-(11)-\sqrt{787.4}}{2*-4.9}=\frac{-11-\sqrt{787.4}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{787.4}}{2*-4.9}=\frac{-11+\sqrt{787.4}}{-9.8}
$