-4.9t^2+22t+0.75=0

Solution for -4.9t^2+22t+0.75=0 equation:

-4.9t^2+22t+0.75=0

a = -4.9; b = 22; c = +0.75;
Δ = b2-4ac
Δ = 222-4·(-4.9)·0.75
Δ = 498.7
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{-(22)-\sqrt{498.7}}{2*-4.9}=\frac{-22-\sqrt{498.7}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+\sqrt{498.7}}{2*-4.9}=\frac{-22+\sqrt{498.7}}{-9.8}
$