4.9t^2+20t-45=0

Solution for 4.9t^2+20t-45=0 equation:

4.9t^2+20t-45=0

a = 4.9; b = 20; c = -45;
Δ = b2-4ac
Δ = 202-4·4.9·(-45)
Δ = 1282
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{-(20)-\sqrt{1282}}{2*4.9}=\frac{-20-\sqrt{1282}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{1282}}{2*4.9}=\frac{-20+\sqrt{1282}}{9.8}
$