-7+2.2t+4.9t^2=0

Solution for -7+2.2t+4.9t^2=0 equation:

-7+2.2t+4.9t^2=0

a = 4.9; b = 2.2; c = -7;
Δ = b2-4ac
Δ = 2.22-4·4.9·(-7)
Δ = 142.04
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{-(2.2)-\sqrt{142.04}}{2*4.9}=\frac{-2.2-\sqrt{142.04}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.2)+\sqrt{142.04}}{2*4.9}=\frac{-2.2+\sqrt{142.04}}{9.8}
$