4.9t^2+22t-140=0

Solution for 4.9t^2+22t-140=0 equation:

4.9t^2+22t-140=0

a = 4.9; b = 22; c = -140;
Δ = b2-4ac
Δ = 222-4·4.9·(-140)
Δ = 3228
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3228}=\sqrt{4*807}=\sqrt{4}*\sqrt{807}=2\sqrt{807}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{807}}{2*4.9}=\frac{-22-2\sqrt{807}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{807}}{2*4.9}=\frac{-22+2\sqrt{807}}{9.8}
$