45=3t+4.9t^2

Solution for 45=3t+4.9t^2 equation:

45=3t+4.9t^2

We move all terms to the left:

45-(3t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-3t+45=0

a = -4.9; b = -3; c = +45;
Δ = b2-4ac
Δ = -32-4·(-4.9)·45
Δ = 891
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:

$\sqrt{\Delta}=\sqrt{891}=\sqrt{81*11}=\sqrt{81}*\sqrt{11}=9\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9\sqrt{11}}{2*-4.9}=\frac{3-9\sqrt{11}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9\sqrt{11}}{2*-4.9}=\frac{3+9\sqrt{11}}{-9.8}
$