-4.9t^2+12t+45=0

Solution for -4.9t^2+12t+45=0 equation:

-4.9t^2+12t+45=0

a = -4.9; b = 12; c = +45;
Δ = b2-4ac
Δ = 122-4·(-4.9)·45
Δ = 1026
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:

$\sqrt{\Delta}=\sqrt{1026}=\sqrt{9*114}=\sqrt{9}*\sqrt{114}=3\sqrt{114}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-3\sqrt{114}}{2*-4.9}=\frac{-12-3\sqrt{114}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+3\sqrt{114}}{2*-4.9}=\frac{-12+3\sqrt{114}}{-9.8}
$