45=0t+0.5(10)t^2

Solution for 45=0t+0.5(10)t^2 equation:

45=0t+0.5(10)t^2

We move all terms to the left:

45-(0t+0.5(10)t^2)=0

We get rid of parentheses

-0.510t^2-0t+45=0

We add all the numbers together, and all the variables

-0.51t^2-1t+45=0

a = -0.51; b = -1; c = +45;
Δ = b2-4ac
Δ = -12-4·(-0.51)·45
Δ = 92.8
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{-(-1)-\sqrt{92.8}}{2*-0.51}=\frac{1-\sqrt{92.8}}{-1.02}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{92.8}}{2*-0.51}=\frac{1+\sqrt{92.8}}{-1.02}
$