4,9t^2+10t-45=0

Solution for 4,9t^2+10t-45=0 equation:

4.9t^2+10t-45=0

a = 4.9; b = 10; c = -45;
Δ = b2-4ac
Δ = 102-4·4.9·(-45)
Δ = 982
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:

$\sqrt{\Delta}=\sqrt{982}=\sqrt{1*982}=\sqrt{1}*\sqrt{982}=1\sqrt{982}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-1\sqrt{982}}{2*4.9}=\frac{-10-1\sqrt{982}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+1\sqrt{982}}{2*4.9}=\frac{-10+1\sqrt{982}}{9.8}
$